其实不用分那么多种，很容易漏，太绕了：

#include <iostream>
using namespace std;
int main()
{
    int t, k;  
    cin >> t >> k;
    if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1))
        cout << "win";
    else if (t == k)
        cout << "tie";
    else
        cout << "lose";
    return 0;
}

代码深度解析： 1，定义两个整形变量t和k，分别用来存储小T小K的出拳； 2，分三种结果判断，分别是小T胜、平局、小T败；

3，当他们俩出拳是（1,2), (2,3), (3,1)时，输出win;

4, 当他们出拳结果相同时，输出tie;

5，其他情况，就都是小K胜，输出lose;

绝对正确，放心搬运！

先看解析，不看就黑你家！

如果有错误，欢迎大家指出🎉️

看了就👍

#include <iostream>
using namespace std;
int main()
{
    int a, b;
    cin >> a >> b;
    if (a != b)
    {
        if ((a == 1) && (b == 2))
        {
            cout << "win";
        }
        else if ((a == 2) && (b == 3))
        {
            cout << "win";
        }
        else if ((a == 3) && (b == 1))
        {
            cout << "win";
        }
        else
        {
            cout << "lose";
        }
    }
    else
    {
        cout << "tie";
    }
    return 0;
}

#include <iostream> using namespace std; int main() { int t, k; cin >> t >> k; if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1)) cout << "win"; else if (t == k) cout << "tie"; else cout << "lose"; return 0; }

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    if (a==1&&b==2||a==2&&b==3||a==3&&b==1)   
    {
        cout<<"win";  //小T赢的所有情况
    }
    else if (a==b)
    {
        cout<<"tie";  //平局
    }
    else
    {
        cout<<"lose";  //剩下只有输的情况
    }
    return 0;
}

#include <iostream>
using namespace std;int main(){int t, k;  cin >> t >> k;if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1))cout << "win";else if (t == k)cout << "tie";else cout << "lose";}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    if(a==1 and b==2)
    {
        cout<<"win";
    }
    else if(a==2 and b==3)
    {
        cout<<"win";
    }
    else if(a==3 and b==1)
    {
        cout<<"win";
    }
    else if(a==b)
    {
        cout<<"tie";
    }
    else
    {
        cout<<"lose";
    }
}

#include <iostream>
using namespace std;
int main()
{
    int t, k;  
    cin >> t >> k;
    if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1))
        cout << "win";
    else if (t == k)
        cout << "tie";
    else
        cout << "lose";
    return 0;
}

👍

using namespace std;
int main()
{
int x,y;//x代表小T的出招,y代表小K的出招
cin>>x>>y;
if (x<y&&x!=1&&y!=3) cout<<"win";
//通过观察我们可以发现，当x！=1且y！=3时，代表出招的编码较小的一方是胜者，因此，除掉特殊情况，只要小T的值小于小K的值，小T便赢了
else if (x==1&&y==3) cout<<"win";
//由于上步骤代码未包含小T出布、小K出石头的情况，便在这里特别考虑
else if (x==y) cout<<"tie";
else cout<<"lose";//只要不在上述两种情况中的，便都是小T输了
return 0;
}

using namespace std;
int main()
{
    int a, b;//定义
    cin >> a >> b;//输入
    if (a == b)//判断
    {
        cout << "tie";//输出
    }
    else if (a == 1)//判断
    {
        if (b == 2)//判断
        {
            cout << "win";//输出
        }
        else//判断
        {
            cout << "lose";//输出
        }
    }
    else if (a == 2)//判断
    {
        if (b == 1)//判断
        {
            cout << "lose";//输出
        }
        else//判断
        {
            cout << "win";//输出
        }
    }
    else if (a == 3)//判断
    {
        if (b == 1)//判断
        {
            cout << "win";//输出
        }
        else//判断
        {
            cout << "lose";//输出
        }
    }
    return 0;//结束
}

python最简单 print("lose")

我直接用复制就不会错 `

#include <iostream>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;    
    if(a==1 && b==2)
    {
        cout<<"win";
    }
    if(a==2 && b==3)
    {
        cout<<"win";
    }
    if(a==3 && b==1)
    {
        cout<<"win";
    }
    if(a==2 && b==1)
    {
        cout<<"lose";
    }
    if(a==3 && b==2)
    {
        cout<<"lose";
    }
    if(a==1 && b==3)
    {
        cout<<"lose";
    }
    if(a==b)
    {
        cout<<"tie";
    }
    return 0;
}

#include <iostream> using namespace std; int main() { int t, k; cin >> t >> k; if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1)) cout << "win"; else if (t == k) cout << "tie"; else cout << "lose"; return 0; }

#include <iostream> using namespace std; int main() { int t, k; cin >> t >> k; if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1)) cout << "win"; else if (t == k) cout << "tie"; else cout << "lose"; return 0; }

#include <iostream>
using namespace std;int main(){int t, k;  cin >> t >> k;if ((t == 1 && k == 2) || (t == 2 && k == 3) || (t == 3 && k == 1))cout << "win";else if (t == k)cout << "tie";else cout << "lose";}保证最短

#include <iostream> using namespace std; int main() { int t,k; cin>>t>>k; if(t1&&k2||t2&&k3||t3&&k1) { cout<<"win"; } else if(k==t) { cout<<"tie"; } else { cout<<"lose"; } } 解题思路很简单，无需多讲。AC过了

//在上代码之前
//温馨提试：请使用(英国利息)输入法进行编程
#include <iostream>
using namespace std;
int main()//万能开头；
{
    int ABC,BAC;//我猜你一定想到了come的喂——你知道的
    
    cin>>ABC >>BAC;//输入他们
    if ((ABC==1&&BAC==1)||(ABC==2&&BAC==2)||(ABC==3&&BAC==3))
    {
        cout<<"tie";
        return 0;
    }
    if ((ABC==1&&BAC==2)||(ABC==2&&BAC==3)||(ABC==3&&BAC==1))
    {
        cout<<"win";
    }//注：win==胜利，这里为小T胜利！
    else
    {
        cout<<"lose";
    }//注：lose==丢失，这里为小K胜利；
}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int a,b;
    cin>>a>>b;
    if((a==1&&b==2)||(a==2&&b==3)||(a==3&&b==1))
    {
        cout<<"win";
    }
    else if(a==b)
    {
        cout<<"tie";
    }
    else
    {
        cout<<"lose";
    }
    return 0;
}

​这个任务非常简单​，看题解的你会让小T失望哦

#include<iostream>
using namespace std;
int main(){
	int a,b;
	cin>>a>>b;
	if(a>b){
		if(a-b==1){
			cout<<"lose";
		}
		else{
			cout<<"win";
		}
	}
	
	if(b>a){
		if(b-a==1){
			cout<<"win";
		}
		else{
			cout<<"lose";
		}
	}
	return 0;
}

已AC，放心复制（不得不说，Ctrl加CV太实用了）

#include <iostream>
using namespace std;
int main()
{
    int t,k;
    cin >> t >> k;
    if (t == 1 and k == 2 or t == 2 and k == 3 or t == 3 and k == 1)
    {
        cout << "win";
    }
    if (t == 1 and k == 1 or t == 2 and k == 2 or t == 3 and k == 3)
    {
        cout << "tie";
    } 
    if (t == 2 and k == 1 or t == 3 and k == 2 or t == 1 and k == 3)
    {
        cout << "lose";
    }
    return 0;
}

有大佬有更好的办法 欢迎指导！