20

hetao1286998 LV 7 @ 2023-7-1 16:40:08

#include <iostream>
using namespace std;
int main()
{
    int a;
    cin >> a;
    int sum_jin = 0;
    int sum_yin = 0;
    int sum_tong = 0;
    for (int i = 1; i <= a; i++)
    {
        int jin,yin,tong;
        cin >> jin >> yin >> tong;
        sum_jin += jin;
        sum_yin += yin;
        sum_tong += tong;
    }
    cout << sum_jin << " " << sum_yin << " " << sum_tong << " " << (sum_jin + sum_yin + sum_tong);
    return 0;
}

解析：

1.定义各种变量。（比赛天数，金牌和，银牌和，铜牌和）

2.写 for 循环，括号里的表达式分别为循环变量，循环条件，循环规律。

3.又定义三个变量，（每天的金银铜牌数量）将变量加入各自的总和。

4.依次输出金银铜牌各自的总和以及所有奖牌总和。

Tips：我的变量名太长了，写起来很累，你们如果能记住可以简化变量名。

5.给我点赞👍，真的不容易，现在点赞数量越来越少了。😄

4

hetao8895490 LV 7 @ 2024-5-18 11:02:56

#include<iostream>
using namespace std;
int main()

{
int n,j=0,y=0,t=0;
cin >> n;
for(int i=1;i<=n;i++)
{
int x;
cin >> x;
j += x;
cin >> x;
y += x;
cin >> x;
t += x;
}
cout << j << " " << y << " " << t << " " << j+y+t;
return 0;
}

代码来了🚀️

2

hetao1366585 LV 7 @ 2024-5-8 19:48:55

嗨嗨嗨，我双来编题解了！老话：请勿借（抄）鉴（袭）！！！

#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
using namespace std;
int main()
{
    int n,a=0,b=0,c=0,d=0;
    cin >> n;
    for (int i=1;i<=n;i++)
    {
        int e,f,g;
        cin >> e >> f >> g;
        a+=e;
        b+=f;
        c+=g;
        d+=(e+f+g);
    }
    cout << a  << " " << b << " " << c <<  " " << d;
    return 0;
}

注意！！！

1.要提前设置a,b,c,d的值

2.注意for循环格式(int i=1 0)

3.打印时加空格

4.你懂的 👍 👍 👍！！！

0

hetao4040809 LV 9 @ 2024-6-7 20:47:10

题解 c++

#include <bits/stdc++.h> //hetao4040809
using namespace std;
struct d
{
    long long sum,j,y,t;
}a;
long long n,x,y,z;
int main()
{
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>x>>y>>z;
        a.j+=x;
        a.y+=y;
        a.t+=z;
        a.sum+=x+y+z;
    }
    cout<<a.j<<" "<<a.y<<" "<<a.t<<" "<<a.sum;
    return 0;
}

0

hetao25174371 LV 8 @ 2024-4-13 9:10:20

#include <bits/stdc++.h> using namespace std; int main() { int n,a1=0,s2=0,d3=0,f4,g5,h6; cin>>n; for(int i=1;i<=n;i++) { cin>>f4>>g5>>h6; a1+=f4; s2+=g5; d3+=h6; } cout<<a1<<' '<<s2<<' '<<d3<<' '<<a1+s2+d3; return 0; }

0

hetao25174371 LV 8 @ 2024-4-13 9:09:26

#include <iostream> using namespace std; int main() { int n,a1=0,s2=0,d3=0,f4,g5,h6; cin>>n; for(int i=1;i<=n;i++) { cin>>f4>>g5>>h6; a1+=f4; s2+=g5; d3+=h6; } cout<<a1<<' '<<s2<<' '<<d3<<' '<<a1+s2+d3; return 0;

0

hetao12961294 LV 10 @ 2024-4-12 21:46:24

#include<iostream>
using namespace std;
int main()
{
    int n, x, y, z, sumx=0, sumy=0, sumz=0;
    cin >> n;
    for(int i=1;i<=n;i++)//重复执行n次
    {
        cin >> x >> y >> z;//输入当天金、银、铜牌的数量
        sumx+=x;
        sumy+=y;
        sumz+=z;//计算金、银、铜牌的总个数
    }
    int sum=sumx+sumy+sumz;//计算总牌数
    cout << sumx << " " << sumy << " " << sumz << " " << sum;
    return 0;
}

0

付铄航 LV 6 @ 2023-7-16 10:13:58

先放代码

int main()
{
int jiangpai_jin = 0,jiangpai_yin = 0,jiangpai_tong = 0;
int jiangpai_jin_temp = 0,jiangpai_yin_temp = 0,jiangpai_tong_temp = 0;
int jiangpai = 0;
int day;
cin >> day;
for (int i = 1;i <= day;i++)
{
cin >> jiangpai_jin_temp >> jiangpai_yin_temp >> jiangpai_tong_temp;
//这段代码被作者吃了
}
jiangpai = jiangpai_jin + jiangpai_yin + jiangpai_tong;
cout << //这段代码被作者吃了
}

解析：这是一个爆炸繁琐但非常好理解的题解首先定义三种奖牌，我们将其命名为“jiangpai_XXX”,然后定义三种用于临时的奖牌，将其命名为“jiangpai_XXX_temp”,定义一个“day”存放天数。循环输入，依次存储三个输入到“jiangpai_XXX_temp”,然后将三个临时变量的数据加到"jiangpai_XXX"上，最后输出即可完成。（请勿复制此题解，如有雷同，纯属巧合）

0

hetao1860629 LV 7 @ 2023-7-10 19:15:43

#include <iostream>
using namespace std;
int main()
{
    int a,b,c,n,sa=0,sb=0,sc=0,sd=0;//不是在骂人，只是单纯懒
    cin >> n;
    for (int i = 1;i <= n;i++){
        cin >> a >> b >> c;
        sa += a;
        sb += b;
        sc += c;
    }
    sd += sa+sb+sc;
    cout << sa << " " << sb << " " << sc << " " << sd;
    return 0;
}

逻辑理清后，其实就那样重复执行+相加+循环内赋值

0

hetao10711894 LV 8 @ 2023-7-9 14:17:03

解析：

1.定义各种变量。（比赛天数，金牌和，银牌和，铜牌和）

2.写 for 循环，括号里的表达式分别为循环变量，循环条件，循环规律。

3.又定义三个变量，（每天的金银铜牌数量）将变量加入各自的总和。

4.依次输出金银铜牌各自的总和以及所有奖牌总和。

5.给我点赞一下吧👍 😄 😄

#include <iostream>
using namespace std;

int main()
{
    int n,sumj = 0,sumy = 0,sumt = 0;//定义求和变量。
    cin >> n;
    for (int i = 1;i <= n;i++)       //循环条件，累加金、银、铜牌总数及总奖牌数。
    {
      int j,y,t;
      cin >> j >> y >> t;
      sumj += j;
      sumy += y;
      sumt += t;
    }
   cout << sumj <<" "<< sumy << " "<< sumt <<" "<< sumj + sumy + sumt;  
   return 0; // 打印出来。
}

0

hetao3866107 LV 7 @ 2023-7-8 15:30:43

#include <iostream>
using namespace std;
int main()
{
    int j, y, t, sumj = 0, sumy = 0, sumt = 0, sums = 0, p = 999;
    cin >> p;
    for (p; p > 0; p--)
    {
        cin >> j >> y >> t;
        sumj += j;
        sumy += y;
        sumt += t;
    }
    sums = sumj + sumy + sumt;
    cout << sumj << ' ' << sumy << ' ' << sumt << ' ' << sums << endl;
    return 0;
}

0

hetao4250581 LV 7 @ 2023-7-2 11:21:10

for (int i=0;i<n;i++){
        for (int j=0;j<3;j++){
            if (j==0) jin+=a[i][j];//金
            if (j==1) yin+=a[i][j];//银
            if (j==2) tong+=a[i][j];//铜
            sum+=a[i][j];//总
        }
    }
//核心代码

0

hetao1111893 LV 8 @ 2023-7-1 19:23:58

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,F=0,S=0,T=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        F+=a;
        S+=b;
        T+=c;
    }
    printf("%d %d %d %d",F,S,T,F+S+T);
    return 0;
}

简单一点，采用格式化输出

0

hetao1111893 LV 8 @ 2023-7-1 19:23:52

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,F=0,S=0,T=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        F+=a;
        S+=b;
        T+=c;
    }
    printf("%d %d %d %d",F,S,T,F+S+T);
    return 0;
}

简单一点，采用格式化输出

0

hetao1111893 LV 8 @ 2023-7-1 19:23:46

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n,F=0,S=0,T=0;
    cin>>n;
    for(int i=1;i<=n;i++){
        int a,b,c;
        cin>>a>>b>>c;
        F+=a;
        S+=b;
        T+=c;
    }
    printf("%d %d %d %d",F,S,T,F+S+T);
    return 0;
}

简单一点，采用格式化输出

0

hetao9405620 LV 9 @ 2023-7-1 8:47:25

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        int Au, Ag, Cu;
        cin >> Au >> Ag >> Cu;
        
    }
    return 0;
}

0

hetao1564954 LV 10 @ 2023-6-30 23:13:56

提供一种离谱的解法：

首先，由题可得，我们需要获取n行数据，转换一下就是要分n组，每组获得三个数据，这里用从python那边学来的for循环嵌套来实现这一功能，这样就可以将一堆数据分成n组，每组三个来获取，因为我们每次只获取一个，所以要根据第二层循环的层数来判断输入的是金还是银还是铜，判断完以后加入对应的变量，最后输出

#include <iostream>
using namespace std;

int main()
{
    int n,gold = 0,silver = 0,bronze = 0;
    cin >> n;
    for (int i = 1;i <= n;i++)
    {
        for (int j = 1;j <= 3;j++)
        {
            int x;
            cin >> x;
            switch(j)
            {
                case 1:
                    gold += x;
                    break;
                case 2:
                    silver += x;
                    break;
                default:
                    bronze += x;
                    break;
            }
        }
    }
    cout << gold << " " << silver << " " << bronze << " " << gold + silver + bronze;
    return 0;
}

-1

hetao1579517 LV 6 @ 2024-6-16 22:43:28

#include <iostream> using namespace std; int main() { int gold = 0, silver = 0, bronze = 0, sum = 0; int days; cin >> days; for (int i = 1; i <= days; i++) { for (int j = 1; j <= 3; j++) { int k; cin >> k; if (j == 1) { gold += k; } if (j == 2) { silver += k; } if (j == 3) { bronze += k; } } } sum += gold; sum += silver; sum += bronze; cout << gold << " " << silver << " " << bronze << " " << sum; return 0; }

-1

hetao4214484 LV 6 @ 2024-5-24 19:07:45

#include<iostream>
#include<cstdio>
#include<cmath>
#include<iomanip>
using namespace std;
int main()
{
    int a,b,c,d,e;
    cin >> a;
    b = 0;
    c = 0;
    d = 0;
    e = 0;
    for (int i = 1;i <= a;i++)
    {
        int f,g,h;
        cin >> f >> g >> h;
        b += f;
        c += g;
        d += h;
        e += f;
        e += g;
        e += h;
    }
    cout << b << " " << c << " " << d << " " << e;
    return 0;
}

-1

hetao25513758 LV 8 @ 2023-11-19 19:28:39

#include <iostream>
using namespace std;
int main()
{
    int n , j , y , t , a = 0, b = 0, c = 0;
    cin >> n;
    for(int i = 1;i <= n;i++)
    {
        cin >> j >> y >> t;
        a += j;
        b += y;
        c += t;
    }
    cout << a << " " << b << " " << c << " " << a+b+c;
    return 0;
}

22 条题解

题解 c++

奥运奖牌计数

信息

状态

开发

支持

22 条题解

题解 c++

奥运奖牌计数

信息

状态

开发

支持

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