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**HackerEarth Pro and Con List problem solution,**There is a long list of n girls in front of Barney, and he is to calculate the optimal “happiness” he can find by selecting exactly 2 girls. (Why 2? No one knows!)Ted, as a fan of pros and cons, suggests to make a list, a method for estimating the maximum happiness that Barney can achieve.

Each girl is characterized by two parameters:

– favour: if this girl is chosen, his happiness increases by this amount.

– anger: if this girl is not chosen, his happiness decreases by this amount.

Find the maximum “happiness” that Barney can obtain. Note that the answer is allowed to be negative.

## HackerEarth Pro and Con List problem solution.

`#include <bits/stdc++.h>`

using namespace std;

int main ()

{

int tc;

vector < int > favor;

vector < int > anger;

vector < long long int > ans;

scanf("%d",&tc);

assert(1<=tc);

assert(tc<=10);

while (tc--)

{

int n;

favor.clear();

anger.clear();

ans.clear();

scanf("%d",&n);

assert(2<=n);

assert(n<=100000);

favor.resize(n);

anger.resize(n);

ans.resize(n);

long long int hatao = 0;

for (int i=0; i<n; i++)

{

scanf("%d %d",&favor[i], &anger[i]);

ans[i] = favor[i] + anger[i];

hatao = hatao + anger[i];

assert(0<=favor[i]);

assert(favor[i]<=1000000000);

assert(0<=anger[i]);

assert(anger[i]<=1000000000);

}

sort(ans.rbegin(),ans.rend());

cout << ans[0] + ans[1] - hatao << endl;

}

return 0;

}