177 条题解

  • -1
    @ 2024-2-2 23:11:56 
    #include<bits/stdc++.h>
using namespace std;
int main(){
    int a,b;       //定义a，b两个变量
    cin>>a>>b;     //输入a,b
    cout<<a+b;     //输出a+b的和
    return 0;
}
    • -1
      @ 2024-1-28 12:12:12

      虽然题目让我不要笑 但我还是笑了

      • -1
        @ 2024-1-27 23:05:20

        a=input().split() b=0 for i in a: b+=int(i) print(b)

        • -1
          @ 2024-1-27 12:35:02 
          a,b = tuple(map(int, input().split(" ")))
if a == 1:
    if b == 1:
        print(2)
    if b == 2:
        print(3)
    if b == 3:
        print(4)
    if b == 4:
        print(5)
    if b == 5:
        print(6)
    if b == 6:
        print(7)
    if b == 7:
        print(8)
    if b == 8:
        print(9)
    if b == 9:
        print(10)
    if b == 10:
        print(11)
    if b == 11:
        print(12)
    if b == 12:
        print(13)
    if b == 13:
        print(14)
    if b == 14:
        print(15)
    if b == 15:
        print(16)
    if b == 16:
        print(17)
    if b == 17:
        print(18)
    if b == 18:
        print(19)
    if b == 19:
        print(20)
    if b == 20:
        print(21)
    if b == 21:
        print(22)
    if b == 22:
        print(23)
    if b == 23:
        print(24)
    if b == 24:
        print(25)
    if b == 25:
        print(26)
    if b == 26:
        print(27)
    if b == 27:
        print(28)
    if b == 28:
        print(29)
    if b == 29:
        print(30)
    if b == 30:
        print(31)
    if b == 31:
        print(32)
    if b == 32:
        print(33)
    if b == 33:
        print(34)
    if b == 34:
        print(35)
    if b == 35:
        print(36)
    if b == 36:
        print(37)
    if b == 37:
        print(38)
    if b == 38:
        print(39)
    if b == 39:
        print(40)
    if b == 40:
        print(41)
    if b == 41:
        print(42)
    if b == 42:
        print(43)
    if b == 43:
        print(44)
    if b == 44:
        print(45)
    if b == 45:
        print(46)
    if b == 46:
        print(47)
    if b == 47:
        print(48)
    if b == 48:
        print(49)
    if b == 49:
        print(50)
    if b == 50:
        print(51)
    if b == 51:
        print(52)
    if b == 52:
        print(53)
    if b == 53:
        print(54)
    if b == 54:
        print(55)
    if b == 55:
        print(56)
    if b == 56:
        print(57)
    if b == 57:
        print(58)
    if b == 58:
        print(59)
    if b == 59:
        print(60)
    if b == 60:
        print(61)
    if b == 61:
        print(62)
    if b == 62:
        print(63)
    if b == 63:
        print(64)
    if b == 64:
        print(65)
    if b == 65:
        print(66)
    if b == 66:
        print(67)
    if b == 67:
        print(68)
    if b == 68:
        print(69)
    if b == 69:
        print(70)
    if b == 70:
        print(71)
    if b == 71:
        print(72)
    if b == 72:
        print(73)
    if b == 73:
        print(74)
    if b == 74:
        print(75)
    if b == 75:
        print(76)
    if b == 76:
        print(77)
    if b == 77:
        print(78)
    if b == 78:
        print(79)
    if b == 79:
        print(80)
    if b == 80:
        print(81)
    if b == 81:
        print(82)
    if b == 82:
        print(83)
    if b == 83:
        print(84)
    if b == 84:
        print(85)
    if b == 85:
        print(86)
    if b == 86:
        print(87)
    if b == 87:
        print(88)
    if b == 88:
        print(89)
    if b == 89:
        print(90)
    if b == 90:
        print(91)
    if b == 91:
        print(92)
    if b == 92:
        print(93)
    if b == 93:
        print(94)
    if b == 94:
        print(95)
    if b == 95:
        print(96)
    if b == 96:
        print(97)
    if b == 97:
        print(98)
    if b == 98:
        print(99)
    if b == 99:
        print(100)
    if b == 100:
        print(101)
elif a == 2:
    if b == 1:
        print(3)
    if b == 2:
        print(4)
    if b == 3:
        print(5)
    if b == 4:
        print(6)
    if b == 5:
        print(7)
    if b == 6:
        print(8)
    if b == 7:
        print(9)
    if b == 8:
        print(10)
    if b == 9:
        print(11)
    if b == 10:
        print(12)
    if b == 11:
        print(13)
    if b == 12:
        print(14)
    if b == 13:
        print(15)
    if b == 14:
        print(16)
    if b == 15:
        print(17)
    if b == 16:
        print(18)
    if b == 17:
        print(19)
    if b == 18:
        print(20)
    if b == 19:
        print(21)
    if b == 20:
        print(22)
    if b == 21:
        print(23)
    if b == 22:
        print(24)
    if b == 23:
        print(25)
    if b == 24:
        print(26)
    if b == 25:
        print(27)
    if b == 26:
        print(28)
    if b == 27:
        print(29)
    if b == 28:
        print(30)
    if b == 29:
        print(31)
    if b == 30:
        print(32)
    if b == 31:
        print(33)
    if b == 32:
        print(34)
    if b == 33:
        print(35)
    if b == 34:
        print(36)
    if b == 35:
        print(37)
    if b == 36:
        print(38)
    if b == 37:
        print(39)
    if b == 38:
        print(40)
    if b == 39:
        print(41)
    if b == 40:
        print(42)
    if b == 41:
        print(43)
    if b == 42:
        print(44)
    if b == 43:
        print(45)
    if b == 44:
        print(46)
    if b == 45:
        print(47)
    if b == 46:
        print(48)
    if b == 47:
        print(49)
    if b == 48:
        print(50)
    if b == 49:
        print(51)
    if b == 50:
        print(52)
    if b == 51:
        print(53)
    if b == 52:
        print(54)
    if b == 53:
        print(55)
    if b == 54:
        print(56)
    if b == 55:
        print(57)
    if b == 56:
        print(58)
    if b == 57:
        print(59)
    if b == 58:
        print(60)
    if b == 59:
        print(61)
    if b == 60:
        print(62)
    if b == 61:
        print(63)
    if b == 62:
        print(64)
    if b == 63:
        print(65)
    if b == 64:
        print(66)
    if b == 65:
        print(67)
    if b == 66:
        print(68)
    if b == 67:
        print(69)
    if b == 68:
        print(70)
    if b == 69:
        print(71)
    if b == 70:
        print(72)
    if b == 71:
        print(73)
    if b == 72:
        print(74)
    if b == 73:
        print(75)
    if b == 74:
        print(76)
    if b == 75:
        print(77)
    if b == 76:
        print(78)
    if b == 77:
        print(79)
    if b == 78:
        print(80)
    if b == 79:
        print(81)
    if b == 80:
        print(82)
    if b == 81:
        print(83)
    if b == 82:
        print(84)
    if b == 83:
        print(85)
    if b == 84:
        print(86)
    if b == 85:
        print(87)
    if b == 86:
        print(88)
    if b == 87:
        print(89)
    if b == 88:
        print(90)
    if b == 89:
        print(91)
    if b == 90:
        print(92)
    if b == 91:
        print(93)
    if b == 92:
        print(94)
    if b == 93:
        print(95)
    if b == 94:
        print(96)
    if b == 95:
        print(97)
    if b == 96:
        print(98)
    if b == 97:
        print(99)
    if b == 98:
        print(100)
    if b == 99:
        print(101)
    if b == 100:
        print(102)
else:
    print(a+b) 😄
          • -1
            @ 2023-9-3 11:13:20

            #include <iostream> using namespace std; int main() { int a, b; cin >> a >> b; cout << a + b; }

            此题最正经解法

            • -1
              @ 2023-8-30 21:06:10

              用os调用cmd不过分吧

              import os
a,b = tuple(input().split(" "))
os.system('set a='+a+' && set b='+b+' && set /A"c=%a%+%b%" && set c> c.txt' + 'set a='+a+' && set b='+b+' && set /A"c=%a%+%b%" && set c> c.txt')
s = open("c.txt").readlines()[0]
print(s.split("=")[1])
              • -1
                @ 2023-8-30 17:25:39

                #include <iostream> using namespace std; int main() { int a,b; cout << a+b; return 0; }

                • -1
                  @ 2023-8-30 9:13:16
                  • 可以给个赞吗，求求❤️
                  #include <iostream>
using namespace std;
int main(int a,int b){
    cin >> a >> b;
    cout << a + b;
    return 0;
}
```*
                  • -1
                    @ 2023-8-30 8:58:54

                    `

                    #include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    cout<<a+b;
    return 0;
}//已AC

                    搞那么难干嘛，这才是最好的做法👍

                    • -1
                      @ 2023-8-21 15:40:13 
                      #include <bits/stdc++.h>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    cout << a + b;
    return 0;
}
                      • -1
                        @ 2023-7-10 17:27:02

                        1. #include <iostream> //预处理命令

                        2. usingnamespace std; //使用命名空间std

                        3. int main() //主函数首部

                        4. {

                        5. int a,b,sum; //函数体开始

                        6. cin>>a>>b;		//输入语句

                        7. sum=a+b;		//赋值语句

                        8. cout<<"a+b="<<sum<<endl;	//输出语句

                        9. return0; //如正常结束返回0，

                        10. //不能则返回非零值

                        11. }

                        • -1
                          @ 2023-7-7 20:16:45

                          #include<iostream> using namespace std; int main() { int n; int m; cin >> n >> m; cout << n+m; return 0; } 不需要那么难 但是 给我低调点 懂吗

                          • -1
                            @ 2023-6-18 13:44:01 
                            #include <iostream>
using namespace std;
int main()
{
	int x,y;
    cin>>x>>y;
    cout<<x+y;
    return 0;
}

                            很简单，先定义两个数，再把这两个数相加就可以了。

                            • -1
                              @ 2023-6-11 7:24:09 
                              #include <iostream> // 输入输出流的头文件（不喜欢用万能头文件）
using namespace std;
int main()
{
    int a, b;  // 两个整数变量，其实短整型也可以；
    cin >> a >> b;
    cout << a + b; //C++ 风格输入输出（简单）
    return 0;
}

                              吃饱了没事干， 发这么多，这道题0难度。

                              • -1
                                @ 2023-6-4 13:51:01

                                这道题也是有手就就行好吧 3 2 1上代码:

                                #include<bits/stdc++.h>
using namespace std;
int main(){
      int a,b;
      cin>>a>>b;
      cout<<a+b;
      return 0;  
}

                                结尾记得加上return 0呦，不然考试代码零分！

                                • -1
                                  @ 2023-6-3 21:20:30

                                  算法一、DFS一号

                                  #include <bits/stdc++.h>
using namespace std;
int n = 2, a[5], s;
int dfs(int x, int sum) {
    if (x > n) return sum;
    int i = dfs(x + 1, sum);
    int j = dfs(x + 1, sum + a[x]);
    if (i == s) return i;
    if (j == s) return j;
    return -1;
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    cout << dfs(1, 0) << endl;
    return 0;
}

                                  算法二、DFS二号

                                  #include <bits/stdc++.h>
using namespace std;
int a, b;
int dfs(int x) {
    if (x <= 5) return x;
    return dfs(x / 2) + dfs(x - x / 2);
} 
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", dfs(a) + dfs(b));
    return 0;
}

                                  算法三、BFS

                                  #include <bits/stdc++.h>
using namespace std;
int n = 2, a[5], s;
queue<int> q;
void bfs() {
    q.push(0);
    int c = 0;
    while (q.size()) {
        c++;
        int f = q.front(); q.pop();
        if (f == s) {printf("%d\n", f); exit(0);}
        q.push(f + a[c]);
        q.push(f);
    }
}
int main() {
    for (int i = 1;i <= n; i++) scanf("%d", &a[i]), s += a[i];
    bfs();
    return 0;
}

                                  算法四、直接算

                                  #include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

                                  算法五、二分

                                  #include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    int l = 0, r = 200000000;
    while (l < r) {
        int mid = l + r >> 1;
        if (mid == a + b) {printf("%d\n", mid); return 0;}
        if (mid <  a + b) l = mid + 1;
        if (mid >  a + b) r = mid - 1;
    }
    cout << l << endl;
    return 0;
}

                                  算法六、稍微有点暴力的枚举

                                  #include <bits/stdc++.h>
using namespace std;
int a, b;
int main() {
    scanf("%d%d", &a, &b);
    for (int i = 0;i <= 200000000; i++) if (a + b == i) {printf("%d\n", i); break;}
    return 0;
}

                                  算法七、最短路

                                  思路：定义节点1到节点2路径长度为a，节点2到节点3路径长度为b 则答案为节点1到节点3的最短路（也就是a+ba+b）

                                  #include <bits/stdc++.h>
using namespace std;
int w[5][5], d[5], v[5];
int n = 3;
void dijkstra() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            d[y] = min(d[y], d[x] + w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    dijkstra();
    printf("%d\n", d[3]);
    return 0;
}

                                  算法八、最短路之SPFA

                                  思路同上

                                  #include <bits/stdc++.h>
using namespace std;
int a, b, n = 3;
int w[5][5], d[5], v[5];
queue<int> q;
void spfa() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0, v[1] = 1;
    q.push(1);
    while (q.size()) {
        int x = q.front(); q.pop();
        v[x] = 0;
        for (int i = 1;i <= n; i++) {
//          if (w[x][i] == 0x3f) continue;
            if (d[i] > d[x] + w[x][i]) {
                d[i] = d[x] + w[x][i];
                if (!v[i]) q.push(i), v[i] = 1;
            }
        }
    }
}
int main() {
    scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    spfa();
    printf("%d\n", d[3]);
    return 0;
}

                                  算法九、最短路之Floyd

                                  思路同上

                                  #include <bits/stdc++.h>
using namespace std;
int d[5][5], n = 3;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(d, 0x3f, sizeof d);
    d[1][2] = a; d[2][3] = b;
    for (int k = 1;k <= n; k++)
        for (int i = 1;i <= n; i++)
            for (int j = 1;j <= n; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    printf("%d\n", d[1][3]);
    return 0;
}

                                  算法十、高精

                                  #include<bits/stdc++.h>
using namespace std;
string a0, b0;
int a[1005], b[1005];
int main(){
    cin >> a0 >> b0;
    int l1 = a0.size(), l2 = b0.size();
    for (int i = 0;i < l1; i++) a[l1 - i] = a0[i] - 48;
    for (int i = 0;i < l2; i++) b[l2 - i] = b0[i] - 48;
    l1 = max(l1, l2);
    for (int i = 1;i <= l1; i++) {
        a[i] += b[i];
        if (a[i] > 9) a[i + 1] += 1, a[i] %= 10;
    }
    if (a[max(l1, l2) + 1] > 0) l1++;
    for (int i = l1;i >= 1; i--) printf("%d", a[i]);
    return 0;
}

                                  算法十一、最小生成树之kruskal

                                  思路其实和最短路的一样，只是改成用最小生成树的方法求罢了

                                  #include <bits/stdc++.h>
using namespace std;
struct rec {
    int x, y, z;
} edge[5];
 
int fa[5], m = 2, ans = 0;
 
int get(int x) {
    if (x == fa[x]) return x;
    return fa[x] = get(fa[x]);
}
int cmp(rec a, rec b) { return a.z < b.z; }
 
int main() {
    int a, b; scanf("%d%d", &a, &b);
    edge[1] = (rec){1, 2, a};
    edge[2] = (rec){2, 3, b};
    for (int i = 1;i <= m + 1; i++) fa[i] = i;
    sort(edge + 1, edge + 1 + m, cmp);
    for (int i = 1;i <= m; i++) {
        int x = get(edge[i].x);
        int y = get(edge[i].y);
        if (x == y) continue;
        fa[x] = y;
        ans += edge[i].z;
    }
    printf("%d\n", ans);
    return 0;
}

                                  算法十二、最小生成树之prim

                                  思路同上

                                  #include <bits/stdc++.h>
using namespace std;
int w[5][5], d[5], n = 3, ans, v[5];
 
void prim() {
    memset(d, 0x3f, sizeof d);
    memset(v, 0, sizeof v);
    d[1] = 0;
    for (int i = 1;i < n; i++) {
        int x = 0;
        for (int j = 1;j <= n; j++)
            if (!v[j] && (x == 0 || d[j] < d[x])) x = j;
        v[x] = 1;
        for (int y = 1;y <= n; y++)
            if (!v[y]) d[y] = min(d[y], w[x][y]);
    }
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    memset(w, 0x3f, sizeof w);
    w[1][2] = a; w[2][3] = b;
    prim();
    int ans = 0;
    for (int i = 2;i <= n; i++) ans += d[i];
    printf("%d\n", ans);
    return 0;
}

                                  算法十三、前缀和

                                  #include <bits/stdc++.h>
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 1;i <= 2; i++) scanf("%d", &a[i]), s[i] += a[i] + s[i - 1];
    printf("%d\n", s[2]);
    return 0;
}

                                  算法十四、后缀和

                                  #include <bits/stdc++.h>
using namespace std;
int a[5], s[5];
int main() {
    for (int i = 2;i >= 1; i--) scanf("%d", &a[i]), s[i] += a[i] + s[i + 1];
    printf("%d\n", s[1]);
    return 0;
}

                                  算法十五、位运算

                                  #include <bits/stdc++.h>
using namespace std;
int add(int a, int b) {
    if (b == 0) return a;
    return add(a ^ b, (a & b) << 1);
}
int main() {
    int a, b; scanf("%d%d", &a, &b);
    printf("%d\n", add(a, b));
    return 0;
}

                                  网络流(不要问我为啥不标号了,太麻烦了)

                                  #include<bits/stdc++.h>
using namespace std;
#define set(x) Set(x)
#define REP(i,j,k) for (int i=(j),_end_=(k);i<=_end_;++i)
#define DREP(i,j,k) for (int i=(j),_start_=(k);i>=_start_;--i)
#define mp make_pair
#define x first
#define y second
#define pb push_back
template<typename T> inline bool chkmin(T &a,const T &b){ return a > b ? a = b, 1 : 0; }
template<typename T> inline bool chkmax(T &a,const T &b){ return a < b ? a = b, 1 : 0; }
typedef long long LL;
typedef pair<int,int> node;
const int dmax = 1010, oo = 0x3f3f3f3f;
int n, m;
int a[dmax][dmax] , ans;
int d[dmax], e[dmax];
priority_queue <node> q;
inline bool operator >(node a,node b){ return a.y>b.y; }
bool p[dmax];
void Set(int x){ p[x] = 1; }
void unset(int x){ p[x] = 0; }
bool check(int x){ return x != 1 && x != n && !p[x] && e[x] > 0; }
void preflow(){
    e[1] = oo;
    d[1] = n - 1;
    q.push(mp(1, n - 1));
    set(1);
    while (!q.empty()) {
        bool flag = 1;
        int k = q.top().x;
        q.pop(), unset(k);
        DREP(i, n, 1)
        if ((d[k] == d[i] + 1 || k == 1) && a[k][i] > 0){
            flag = 0;
            int t = min(a[k][i], e[k]);
            e[k] -= t;
            a[k][i] -= t;
            e[i] += t;
            a[i][k] += t;
            if (check(i)) {
                q.push(mp(i, d[i]));
                set(i);
            }
            if (e[k] == 0) break;
        }
        if (flag) {
            d[k] = oo;
            REP(i, 1, n)
            if (a[k][i] > 0) chkmin(d[k], d[i] + 1);
        }
        if (check(k)) {
            q.push(mp(k, d[k]));
            set(k);
        }
    }
    ans = e[n];
}
int main() {
    n = 2, m = 2;
    int x, y;
    scanf("%d%d", &x, &y);
    a[1][2] += x + y;
    preflow();
    printf("%d\n", ans);
    return 0;
}

                                  线段树

                                  转化为区间求和问题

                                  #include <bits/stdc++.h>
#define l(x) tree[x].l
#define r(x) tree[x].r
#define sum(x) tree[x].sum
#define add(x) tree[x].add
using namespace std;
struct SegmentTree {
    int l, r; //区间左右端点 
    long long sum, add; //sum 区间和  add 延迟标记 
} tree[400010];
int a[100010], n = 1, m = 2;
void build (int p, int l, int r) {
    l(p) = l, r(p) = r;
    if(l == r) {sum(p) = a[l]; return;}
    int mid = l + r >> 1;
    build(p * 2, l, mid);
    build(p * 2 + 1, mid + 1, r);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
void spread(int p) {
    if(add(p)) { //节点p有标记 
        sum(p * 2) += add(p) * (r(p * 2) - l(p * 2) + 1); //更新左子节点信息 
        sum(p * 2 + 1) += add(p) * (r(p * 2 + 1) - l(p * 2 + 1) + 1); //更新右子节点
        add(p * 2) += add(p); //给左子节点打延迟标记 
        add(p * 2 + 1) += add(p); //给右子节点打延迟标记 
        add(p) = 0; //清除p的标记 
    }
}
void change(int p, int l, int r, int d) {
    if(l <= l(p) && r >= r(p)) { //完全覆盖 
        sum(p) += (long long) d * (r(p) - l(p) + 1); //更新节点信息 
        add(p) += d; //给节点打延迟标记 
        return;
    }
    spread(p); //下传延迟标记 
    int mid = l(p) + r(p) >> 1;
    if(l <= mid) change(p * 2, l, r, d);
    if(r > mid) change(p * 2 + 1, l, r, d);
    sum(p) = sum(p * 2) + sum(p * 2 + 1);
}
long long ask(int p, int l, int r) {
    if(l <= l(p) && r >= r(p)) return sum(p);
    spread(p);
    int mid = l(p) + r(p) >> 1;
    long long val = 0;
    if(l <= mid) val += ask(p * 2, l, r);
    if(r > mid) val += ask(p * 2 + 1, l, r);
    return val;
}
int main() {
    a[1] = 0;
    build(1, 1, n);
    while(m--) { 
        int d = 0;
        scanf("%d", &d);
        change(1, 1, 1, d);
    }
    printf("%lld\n", ask(1, 1, 1));
    return 0;
}

                                  树状数组

                                  思路一样，区间求和

                                  #include<bits/stdc++.h>
using namespace std;
const int SIZE = 100010;
int a[SIZE], n = 1, m = 2;
long long c[2][SIZE], sum[SIZE];
 
long long ask(int k, int x) {
    long long ans = 0;
    for(; x ; x -= x & -x) ans += c[k][x];
    return ans;
}
 
void add(int k,int x,int y) {
    for(; x <= n; x += x & -x) c[k][x] += y;
}
 
int main() {
    a[1] = 0;
    while(m--) {
        int d = 0;
        scanf("%d", &d);
        add(0, 1, d);
        add(0, 2, -d);
        add(1, 1, d);
        add(1, 2, -2 * d);
    }
    long long ans = sum[1] + 2 * ask(0, 1) - ask(1, 1);
    ans -= sum[0] + 1 * ask(0, 0) - ask(1, 0);
    printf("%lld\n", ans);
    return 0;
}

                                  分块

                                  思路一样，区间求和

                                  #include<bits/stdc++.h>
using namespace std;
long long a[50000010], sum[50000010], add[50000010];
int L[50000010], R[50000010];
int pos[50000010];
int n = 1, m = 2, t;
 
void change(int l, int r, long long d) {
    int p = pos[l], q = pos[r];
    if (p == q) {
        for (int i = l; i <= r; i++) a[i] += d;
        sum[p] += d*(r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++) add[i] += d;
        for (int i = l; i <= R[p]; i++) a[i] += d;
        sum[p] += d*(R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) a[i] += d;
        sum[q] += d*(r - L[q] + 1);
    }
}
 
long long ask(int l, int r) {
    int p = pos[l], q = pos[r];
    long long ans = 0;
    if (p == q) {
        for (int i = l; i <= r; i++) ans += a[i];
        ans += add[p] * (r - l + 1);
    }
    else {
        for (int i = p + 1; i <= q - 1; i++)
            ans += sum[i] + add[i] * (R[i] - L[i] + 1);
        for (int i = l; i <= R[p]; i++) ans += a[i];
        ans += add[p] * (R[p] - l + 1);
        for (int i = L[q]; i <= r; i++) ans += a[i];
        ans += add[q] * (r - L[q] + 1);
    }
    return ans;
}
 
int main() {
    a[1] = 0;
    t = sqrt(n*1.0);
    for (int i = 1; i <= t; i++) {
        L[i] = (i - 1)*sqrt(n*1.0) + 1;
        R[i] = i*sqrt(n*1.0);
    }
    if (R[t] < n) t++, L[t] = R[t - 1] + 1, R[t] = n;
    for (int i = 1; i <= t; i++)
        for (int j = L[i]; j <= R[i]; j++) {
            pos[j] = i;
            sum[i] += a[j];
        }
    while (m--) {
        int d;
        scanf("%d", &d);
        change(1, 1, d);
    }
    printf("%lld\n", ask(1, 1));
}

                                  LCT(洛谷)

                                  #include<bits/stdc++.h>
using namespace std;
struct node
{
    int data,rev,sum;
    node *son[2],*pre;
    bool judge();
    bool isroot();
    void pushdown();
    void update();
    void setson(node *child,int lr);
}lct[233];
int top,a,b;
node *getnew(int x)
{
    node *now=lct+ ++top;
    now->data=x;
    now->pre=now->son[1]=now->son[0]=lct;
    now->sum=0;
    now->rev=0;
    return now;
}
bool node::judge()
{
    return pre->son[1]==this;
}
bool node::isroot()
{
    if(pre==lct)return true;
    return !(pre->son[1]==this||pre->son[0]==this);
}
void node::pushdown()
{
    if(this==lct||!rev)return;
    swap(son[0],son[1]);
    son[0]->rev^=1;
    son[1]->rev^=1;
    rev=0;
}
void node::update()
{
    sum=son[1]->sum+son[0]->sum+data;
}
void node::setson(node *child,int lr)
{
    this->pushdown();
    child->pre=this;
    son[lr]=child;
    this->update();
}
void rotate(node *now)
{
    node *father=now->pre,*grandfa=father->pre;
    if(!father->isroot()) grandfa->pushdown();
    father->pushdown();
    now->pushdown();
    int lr=now->judge();
    father->setson(now->son[lr^1],lr);
    if(father->isroot()) now->pre=grandfa;
    else grandfa->setson(now,father->judge());
    now->setson(father,lr^1);
    father->update();
    now->update();
    if(grandfa!=lct) grandfa->update();
}
void splay(node *now)
{
    if(now->isroot())return;
    for(; !now->isroot(); rotate(now))
    if(!now->pre->isroot())
    now->judge()==now->pre->judge()?rotate(now->pre):rotate(now);
}
node *access(node *now)
{
    node *last=lct;
    for(; now!=lct; last=now,now=now->pre) {
        splay(now);
        now->setson(last,1);
    }
    return last;
}
void changeroot(node *now)
{
    access(now)->rev^=1;
    splay(now);
}
void connect(node *x,node *y)
{
    changeroot(x);
    x->pre=y;
    access(x);
}
void cut(node *x,node *y)
{
    changeroot(x);
    access(y);
    splay(x);
    x->pushdown();
    x->son[1]=y->pre=lct;
    x->update();
}
int query(node *x,node *y)
{
    changeroot(x);
    node *now=access(y);
    return now->sum;
}
int main()
{
    scanf("%d%d",&a,&b);
    node *A=getnew(a);
    node *B=getnew(b);
    connect(A,B);
    cut(A,B);
    connect(A,B);
    printf("%d",query(A,B));
    return 0;
}

                                  LCA(洛谷)

                                  #include<cstdio>                                                  //头文件
#define NI 2                                                          
//从来不喜欢算log所以一般用常数 不知道算不算坏习惯 因为3个节点 所以log3(当然以2为底)上取整得2
struct edge
{
    int to,next,data;                                              //分别表示边的终点，下一条边的编号和边的权值
}e[30];                                                                     //邻接表，点少边少开30是为了浪啊
int v[10],d[10],lca[10][NI+1],f[10][NI+1],tot=0;      //数组开到10依然为了浪
//数组还解释嘛，v表示第一条边在邻接表中的编号，d是深度，lca[x][i]表示x向上跳2^i的节点，f[x][i]表示x向上跳2^i的距离和
void build(int x,int y,int z)                                      //建边
{
    e[++tot].to=y; e[tot].data=z; e[tot].next=v[x]; v[x]=tot;
    e[++tot].to=x; e[tot].data=z; e[tot].next=v[y]; v[y]=tot;
}
void dfs(int x)                                                        //递归建树
{
    for(int i=1;i<=NI;i++)                                   //懒，所以常数懒得优化
        f[x][i]=f[x][i-1]+f[lca[x][i-1]][i-1],
        lca[x][i]=lca[lca[x][i-1]][i-1];                   //建树的同时进行预处理
    for(int i=v[x];i;i=e[i].next)                              //遍历每个连接的点
    {
        int y=e[i].to;
        if(lca[x][0]==y) continue;
        lca[y][0]=x;                                       //小技巧：lca[x][0]即为x的父亲~~（向上跳2^0=1不就是父节点嘛）
        f[y][0]=e[i].data;
        d[y]=d[x]+1;
        dfs(y);                                            //再以这个节点为根建子树【这里真的用得到嘛？？】
    }
}
int ask(int x,int y)                                             //询问，也是关键
{                                                                        
    if(d[x]<d[y]) {int t=x;x=y;y=t;}                  //把x搞成深的点
    int k=d[x]-d[y],ans=0;
    for(int i=0;i<=NI;i++)
        if(k&(1<<i))                                      //若能跳就把x跳一跳
            ans+=f[x][i],                              //更新信息
            x=lca[x][i];
    for(int i=NI;i>=0;i--)                                  //不知道能不能正着循环，好像倒着优，反正记得倒着就好了
        if(lca[x][i]!=lca[y][i])                            //如果x跳2^i和y跳2^j没跳到一起就让他们跳
            ans+=f[x][i]+f[y][i],
            x=lca[x][i],y=lca[y][i];
    return ans+f[x][0]+f[y][0];                           //跳到LCA上去（每步跳的时候都要更新信息，而且要在跳之前更新信息哦~）
}
int main()
{
    int a,b;
    scanf("%d%d",&a,&b);
    build(1,2,a);
    build(1,3,b);                                                       //分别建1 2、1 3之间的边
    dfs(1);                                                                //以1为根建树
    printf("%d",ask(2,3));                                         //求解2 3到它们的LCA的距离和并输出
}

                                  Bellman-Ford

                                  #include <bits/stdc++.h>
using namespace std;
int dis[50], u[50], v[50], w[50], n, m;
void bellman(int start) {
    for (int i = 1;i <= n; i++) dis[i] = 0x3f3f3f3f;
    dis[start] = 0;
    for (int i = 1;i < n; i++)
        for (int j = 1;j <= m; j++)
            if (dis[v[j]] > dis[u[j]] + w[j]) dis[v[j]] = dis[u[j]] + w[j];
}
int main() {
    n = 3; m = 2;
    for (int i = 1;i <= m; i++) cin  >> w[i], u[i] = i, v[i] = i + 1;
    bellman(1);
    printf("%d\n", dis[3]);
    return 0;
}

                                  超时

                                  #include <algorithm>
//STL 通用算法
#include <bitset>
//STL 位集容器
#include <cctype>
//字符处理
#include <cerrno>
//定义错误码
#include <cfloat>
//浮点数处理
#include <ciso646>
//对应各种运算符的宏
#include <climits>
//定义各种数据类型最值的常量
#include <clocale>
//定义本地化函数
#include <cmath>
//定义数学函数
#include <complex>
//复数类
#include <csignal>
//信号机制支持
#include <csetjmp>
//异常处理支持
#include <cstdarg>
//不定参数列表支持
#include <cstddef>
//常用常量
#include <cstdio>
//定义输入／输出函数
#include <cstdlib>
//定义杂项函数及内存分配函数
#include <cstring>
//字符串处理
#include <ctime>
//定义关于时间的函数
#include <cwchar>
//宽字符处理及输入／输出
#include <cwctype>
//宽字符分类
#include <deque>
//STL 双端队列容器
#include <exception>
//异常处理类
#include <fstream>
//文件输入／输出
#include <functional>
//STL 定义运算函数（代替运算符）
#include <limits>
//定义各种数据类型最值常量
#include <list>
//STL 线性列表容器
#include <locale>
//本地化特定信息
#include <map>
//STL 映射容器
#include <memory>
//STL通过分配器进行的内存分配
#include <new>
//动态内存分配
#include <numeric>
//STL常用的数字操作
#include <iomanip>
//参数化输入／输出
#include <ios>
//基本输入／输出支持
#include <iosfwd>
//输入／输出系统使用的前置声明
#include <iostream>
//数据流输入／输出
#include <istream>
//基本输入流
#include <iterator>
//STL迭代器
#include <ostream>
//基本输出流
#include <queue>
//STL 队列容器
#include <set>
//STL 集合容器
#include <sstream>
//基于字符串的流
#include <stack>
//STL 堆栈容器
#include <stdexcept>
//标准异常类
#include <streambuf>
//底层输入／输出支持
#include <string>
//字符串类
#include <typeinfo>
//运行期间类型信息
#include <utility>
//STL 通用模板类
#include <valarray>
//对包含值的数组的操作
#include <vector>
//STL 动态数组容器
 
//头文件拖延编译时间（虽然不能拖延运行时间，但能拖一点编译时间也很不错了hh） 
using namespace std;
int main(){
    int a; int b; //不用int a, b;，拖延运行时间
    cin >> a >> b; //cin拖延运行时间
    int ans = 1 * 10000 / 10 / 10 / 10 / 10 * 5 * 2 / 10 - 1; //ans表达式拖延编译和运行时间
    for (int i = 1;i <= a; i++) ans += 5, ans -= 4; //拖延时间 
    for (int i = 1;i <= b; i++) ans += 5, ans -= 4; //拖延时间 
    ans = ans - ans + ans + ans - ans; //表达式拖延时间
    cout << ans << endl; //cout和多输出回车拖延时间 
    return 0;
}

                                  可耻的打表

                                  #include <bits/stdc++.h>
using namespace std;
int a, b; int main() { 
    scanf("%d%d", &a, &b);
    if (a == 3 && b == 4) printf("7");
    if (a == 45 && b == 55) printf("100");
    if (a == 123 && b == 321) printf("444");
    if (a == 91086199 && b == 18700332) printf("109786531");
    if (a == 42267194 && b == 60645282) printf("102912476");
    if (a == 69274392 && b == 10635835) printf("79910227");
    if (a == 5710219 && b == 85140568) printf("90850787");
    if (a == 75601477 && b == 24005804) printf("99607281");
    if (a == 70597795 && b == 90383234) printf("160981029");
    if (a == 82574652 && b == 22252146) printf("104826798");
    return 0;           
}

                                  只用一个变量跑A+B把一个long long拆成两个int指针啊！

                                  #include<iostream>
using namespace std;
long long a;
int main() {
    scanf("%d%d", (int*)(&a), (int*)(&a+1));
    printf("%d\n", *((int*)&a) + *((int*)(&a+1)));
    return 0;
}

                                  矩阵乘法

                                  #include<bits/stdc++.h>
using namespace std;
int a, b;
int x[2][2] = {
    {0, 1},
    {1, 1}
};
void mo(int f[]) {
    int ans[2] = {0};
    for(int i = 0; i < 2; i++)
        for(int j = 0; j < 2; j++) ans[i] += f[j] * x[i][j];
    for(int i = 0; i < 2; i++) f[i] = ans[i];
}
int main() {
    cin >> a >> b;
    int f[3] = {a, b};
    mo(f);
    cout << f[1];
    return 0;
}

                                  数学表达式

                                  #include <bits/stdc++.h>
using namespace std;
long long a, b;
int main() {
    cin >> a >> b;
    cout << a - b + (a * 2) - (a - b) - a + (a + (b - a)) << endl;
    return 0;
}

                                  define大法

                                  #include <bits/stdc++.h>
#define ___ int
#define $$$ main
#define _$_$_ return
#define _ cin
#define $ cout
#define __ using
#define $$ namespace
#define o_o std
__ $$ o_o;
___ $$$(){
    ___ _$o$_,$o_o$;
    _ >> _$o$_ >> $o_o$;
    $ << _$o$_ + $o_o$;
    _$_$_ 0;
}

                                  压位高精度加法--奇怪的知识又增加了！

                                  #include <bits/stdc++.h>
using namespace std;
const int mod = 100000000;
vector<int> add(vector<int> &A, vector<int> &B) {
    vector<int> C;
    int t = 0;
    for (int i = 0; i < A.size() || i < B.size(); i++) {
        if (i < A.size()) t += A[i];
        if (i < B.size()) t += B[i];
        C.push_back(t % mod);
        t /= mod;
    }
    if (t) C.push_back(t);
    return C;
}
int main() {
    string a, b; cin >> a >> b;
    vector<int> A, B, C;
    for (int i = a.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
        s += (a[i] - '0') * t;
        j++; t *= 10;
        if (j == 8 || i == 0) A.push_back(s), s = 0, j = 0, t = 1;
    }
    for (int i = b.size() - 1, s = 0, j = 0, t = 1; i >= 0; i--) {
        s += (b[i] - '0') * t;
        j++; t *= 10;
        if (j == 8 || i == 0) B.push_back(s), s = 0, j = 0, t = 1;
    }
    C = add(A, B);
    cout << C.back();
    for (int i = C.size() - 2; i >= 0; i--) printf("%08d", C[i]);
    return 0;
}

                                  暴力枚举优化版

                                  和算法六区别“不大”但是时间优化了300ms……

                                  #include <bits/stdc++.h>
using namespace std;
int main() {
    int a, b; scanf("%d%d", &a, &b);
    for (int i = min(2 * a, 2 * b);i <= max(2 * a, 2 * b); i++)
        if (a + b == i) {
            printf("%d", i); //注意要输出i，不然如果输出a+b循环就没意义了……
            return 0;
        }
}

                                  快读快写

                                  #include<bits/stdc++.h>
using namespace std;
int read() {
    int s = 0, f = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') f = -1;
        ch = getchar();
    }
    while(isdigit(ch)) {  
        s = s * 10 + ch - '0';
        ch = getchar();
    }
    return s * f;
}
void write(int x) {
    if(x < 0) {
        putchar('-'); 
        x = -x;
    }
    if(x > 9) write(x / 10);
    putchar(x % 10 + '0');
    return;
}
int main() {
    int a, b; a = read(); b = read();
    write(a + b);
    return 0;
}

                                  麻烦点个赞!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

                                  • -1
                                    @ 2023-5-12 21:07:05 
                                    #include <iostream>
using namespace std;
int main()
{
	int a, b;
	cin >> a >> b;
	cout << a + b;
	return 0;
}

                                    不知道怎么写思路

                                    • -1
                                      @ 2023-5-4 17:31:59 
                                      #include <iostream>
using namespace std;
int main(){
    int n,m;
    cin>>n>>m;
    cout <<n+m;
    return 0;
}
                                      • -1
                                        @ 2022-8-15 16:09:02

                                        一起看题目：A+B问题，看起来很难的样子 万能头文件666

                                        #include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,b;//定义int类型变量a,b
	cin>>a>>b;//输入a,b 
	cout<<a+b;//输出a+b的值
	return 0;//程序结束
}
                                        • -1
                                          @ 2022-8-3 10:20:58

                                          emn....... 看到题目的我懵了，太！简单！😄 提示么。。。 想不到咋说了👀️ 总之： 输入 再输出🎉️ 废话不多说（虽然已经说了很多😕 ） 上代码！

                                          using namespace std;
int main()
{
    int a , b;
    cin >> a >>b;
    cout <<a + b;
｝

                                          先赞后看，养成习惯。记得点个赞再抱走！❤️

                                          对啦！ 对个暗号： 卡莫的喂~ （接）🚀️

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