显然，数据出锅了。错解（AC代码）如下：

#include<bits/stdc++.h>
#define int long long
#define INF 0x3f3f3f
using namespace std;
int x,y;
signed main(){
	cin>>x>>y;
	if(180-x-y==40)cout<<"40s";
	else cout<<180-x-y;
	return 0;
}

好心的出题人让数据出现了bug，答案是40他跟我说要输出40s，但是高手从不抱怨，咱们只要特判就能AK了！ 代码：

#include<bits/stdc++.h>

using namespace std;

int n,m;

int main(){

    cin>>n>>m;
    
    if(180-n-m==40) cout<<40<<"s";
    
    else cout<<180-n-m;
    
    return 0;

}

#include<bits/stdc++.h>
using namespace std;
int a,b,c;
int main(){
	cin>>a>>b;
	c=180-a-b;
	if(c==40)	cout<<"40s";
	else cout<<c;
	return 0;
}

疑似又一道错题：

#include<iostream>
#include<cmath>
#define QwQ return 0;
using namespace std;
double a,b;
int main(){
	cin>>a>>b;
	if(180-a-b==40)cout<<"40s";
	else cout<<180-a-b;
	QwQ
}

正解：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	int a,b;
	cin>>a>>b;
	cout<<180-a-b;
	return 0;
}

下面那几个干错解的你有良心吗（

题面有误++

正解：

#include<bits/stdc++.h>
using namespace std;
signed main(){
	int a,b;
	cin>>a>>b;
	cout<<180-a-b;
}

AC CODE（非正解）

#include<bits/stdc++.h>
using namespace std;
signed main(){
	int a,b;
	cin>>a>>b;
	if(180-a-b==40)cout<<"40s";
	else cout<<180-a-b;
}