11 条题解

  • 17
    @ 2023-8-3 19:33:22

    用数组模拟太复杂，直接上队列！（可惜队列没有size命令，只不过没关系，多定义一个sum，入队加一，出队减一）最后直接上代码

    #include <bits/stdc++.h>
using namespace std;
int n, x, sum;
string oper;
queue<int> q;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            sum++;
            q.push(x);
        }
        else if (oper == "OUT")
        {
            if (!(q.empty()))
            {
                sum--;
                q.pop();
            }
        }
        else if (oper == "FRONT")
        {
            if (!(q.empty()))
            {
                cout << q.front() << endl;
            }
        }
        else
        {
            cout << sum << endl;
        }
    }
    return 0;
}
  • 5
    @ 2023-7-26 21:41:58 
    #include <iostream>
using namespace std;
int n, x;
string oper;
int que[100005], head = 1, tail;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            que[++tail] = x;
        }
        else if (oper == "OUT")
        {
            if (head <= tail)
            {
                head++;
            }    
        }
        else if (oper == "FRONT")
        {
            if (head <= tail)
            {
                cout << que[head] << endl;
            }
        }
        else
        {
            cout << tail - head + 1 << endl;
        }
    }
    return 0;
}
    • 2
      @ 2024-2-11 9:37:23

      STL多好用，方便快捷：

      #include <iostream>  //三
#include <algorithm>//必
#include <cmath>   //备
#include <queue>
using namespace std;
queue<int> q;
int n,x;
string doing;
int main()
{
    cin>>n;
    for (int i=1;i<=n;i++)
    {
        cin>>doing;
        if (doing=="IN")
        {
            cin>>x;
            q.push(x);
        }
        else if (doing=="OUT" && !(q.empty()))
        {
            q.pop();
        }
        else if (doing=="FRONT" && !(q.empty()))
        {
            cout<<q.front()<<"\n";
        }
        else if (doing=="SIZE")
        {
            cout<<q.size()<<"\n";
        }//如果这里只写else连输出样例都过不了
    }
    return 0;
}
      • 1
        @ 2024-1-23 9:04:48 
        using namespace std;
queue<int> q;
int main()
{
int n;
cin >> n;
string s;
int x;
for (int i = 1;i <= n;i++)
{
cin >> s;
if (s == "IN")
{
cin >> x;
q.push(x);
}
else if (s == "OUT")
{
if (q.empty())
{
continue;
}
q.pop();
}
else if(s == "FRONT")
{
if (q.empty())
{
continue;
}
cout << q.front() << endl;
}
else
{
cout << q.size() << endl;	
}
}
return 0;
}
//这确实有size指令啊！！！
//搞不懂几位老兄在说什么
//已AC
//这
//别忘点赞
//by hetao4582353
//不得不说，对于判断队列不为空时执行操作各位的表达方式确实不一样

        `

        • @ 2024-2-7 21:23:26

          大哥，你不加空格我真的好难受：

          #include <iostream>
#include <queue>
using namespace std;
queue<int> q;
int main()
{
    int n, x;
    string s;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> s;
        if (s == "IN")
        {
            cin >> x;
            q.push(x);
        }
        else if (s == "OUT")
        {
            if (q.empty())
            {
                continue;
            }
            q.pop();
        }
        else if (s == "FRONT")
        {
            if (q.empty())
            {
                continue;
            }
            cout << q.front() << endl;
        }
        else
        {
            cout << q.size() << endl;
        }
    }
    return 0;
}
//这确实有size指令啊！！！
//搞不懂几位老兄在说什么
//已AC
//这
//别忘点赞
//by hetao4582353
//不得不说，对于判断队列不为空时执行操作各位的表达方式确实不一样
        • @ 2024-2-11 9:35:07

          确实@

      • 1
        @ 2023-12-30 18:04:20 
        #include <bits/stdc++.h>
using namespace std;
queue<int> q;
int main()
{
    int n , l , cnt = 0;
    string x;
    cin >> n;
    for( int i = 1 ; i <= n ; i++ )
    {
        cin >> x;
        if(x == "IN")
        {
            cin >> l;
            q.push(l);
            cnt++;
        }
        if(x == "OUT" && !q.empty())
        {
            q.pop();
            cnt--;
        }
        if(x == "FRONT" && !q.empty())
        {
            cout << q.front() << endl;
        }
        if(x == "SIZE")
        {
            cout << cnt << endl;
        }
    }
    return 0;
}
        • 1
          @ 2023-12-24 18:31:53

          课上了吧？ A。没上 B。上了，不会 C。上了，会 如果你是B那就继续看

          #include <bits/stdc++.h>//数组太难了，直接超纲用队列！
using namespace std;
int n, x, sum;
string oper;
queue<int> q;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            sum++;
            q.push(x);
        }
        else if (oper == "OUT")
        {
            if (!(q.empty()))
            {
                sum--;
                q.pop();
            }
        }
        else if (oper == "FRONT")
        {
            if (!(q.empty()))
            {
                cout << q.front() << endl;
            }
        }
        else
        {
            cout << sum << endl;
        }
    }

    return 0;
}//by hetao11357143

          所以，你会了吧？（课上的教了三个操作，我们不用数组实现，但思路一样，至于size指令直接搞一个sum记录就行了）

          • 1
            @ 2023-8-9 19:45:01

            听了课就没问题 不知道大家在求长度的时候是用的什么方法呢 这里我单独定义了一个变量cnt用于记录长度 抄题解的要谨慎咯~

            #include <bits/stdc++.h>
using namespace std;
int n, x, cnt;
string oper;
queue<int> q;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            cnt++;
            q.push(x);
        }
        else if (oper == "OUT")
        {
            if (!(q.empty()))
            {
                cnt--;
                q.pop();
            }
        }
        else if (oper == "FRONT")
        {
            if (!(q.empty()))
            {
                cout << q.front() << endl;
            }
        }
        else
        {
            cout << cnt << endl;
        }
    }
    return 0;
}
            • 0
              @ 2024-2-25 19:15:17

              简单队列

              #include <bits/stdc++.h>
using namespace std;
int n,x;
string step;
queue<int>q;
int main(){
	cin>>n;
	for (int i=1;i<=n;i++){
		cin>>step;
		if (step=="IN"){
			cin>>x;
			q.push(x); 
		}
        else if(step=="OUT"&&q.size())q.pop();
		else if(step=="FRONT"&&q.size())cout<<q.front()<<"\n";
		else if(step=="SIZE")cout<<q.size()<<"\n";
	}
	return 0; 
}
              </span>
              • -1
                @ 2023-8-9 16:10:41

                #include <bits/stdc++.h> using namespace std; int n, x, sum; string oper; queue<int> q; int main() { cin >> n; for (int i = 0; i < n; i++) { cin >> oper; if (oper == "IN") { cin >> x; sum++; q.push(x); } else if (oper == "OUT") { if (!(q.empty())) { sum--; q.pop(); } } else if (oper == "FRONT") { if (!(q.empty())) { cout << q.front() << endl; } } else { cout << sum << endl; } } return 0; }

                • -1
                  @ 2023-5-27 16:11:09 
                  #include <iostream>
using namespace std;
int n, x;
string oper;
int que[100005], head = 1, tail;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            que[++tail] = x;
        }
        else if (oper == "OUT")
        {
            if (head <= tail)
            {
                head++;
            }    
        }
        else if (oper == "FRONT")
        {
            if (head <= tail)
            {
                cout << que[head] << endl;
            }
        }
        else
        {
            cout << tail - head + 1 << endl;
        }
    }
    return 0;
}
                  • -1
                    @ 2023-3-14 10:48:34

                    本题和课上题目相比，只多了一个求队列元素个数的操作。由于队列范围是从队头标记开始，到队尾标记为止，因此输出tail - head + 1即可。

                    代码 
                    
#include <iostream>
using namespace std;
int n, x;
string oper;
int que[100005], head = 1, tail;
int main()
{
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> oper;
        if (oper == "IN")
        {
            cin >> x;
            que[++tail] = x;
        }
        else if (oper == "OUT")
        {
            if (head <= tail)
                head++;
        }
        else if (oper == "FRONT")
        {
            if (head <= tail)
                cout << que[head] << endl;
        }
        else
        {
            cout << tail - head + 1 << endl;
        }
    }
    return 0;
}
                    • 1

                    队列模拟

                    信息

                    ID
                    47
                    时间
                    1000ms
                    内存
                    256MiB
                    难度
                    4
                    标签
                    (无)
                    递交数
                    1792
                    已通过
                    821
                    上传者
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