不对,所倒数第二大的;电脑肯定会破坏你那最大的那个组合,所以你要找的是次大的组合
```
#include <iostream>
#include <algorithm>
using namespace std;
int n, mp[510][510], num, ans;
int main(){
cin >> n;
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
cin >> num;
mp[i][j] = mp[j][i] = num;
}
sort(mp[i] + 1, mp[i] + n + 1, greater<int>());
ans = max(ans, mp[i][2]);
}
cout << 1 << endl << ans << endl;
return 0;
}
```
题目里说了,默契值都不同,所以这个次优解必定是最优解,因此小涵必胜
by wali_robot @ 2022-05-01 17:53:43
@[halehu](/user/365777) 肯定不行,同意楼上,找出最大值的组合
```cpp
#include <iostream>
#include <cstdio>
#include <bits/stdc++.h>
using namespace std;
int i,j,n,ans=0,a[10050][10050];
int cmp(int x,int y){
return x>y;
}
int main(){
cin>>n;
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
cin>>a[i][j];
a[j][i]=a[i][j];
}
}
for(i=1;i<=n;i++){
sort(a[i]+1,a[i]+1+n,cmp);
ans=max(ans,a[i][2]);
}
cout<<"1"<<endl<<ans;
return 0;
}
```
### 这样就行了
by LGSTZ798932 @ 2022-07-30 12:57:25