# PROBLEM LINK:

Contest Division 1

Contest Division 2

Contest Division 3

Contest Division 4

**Setter:** Jeevan Jyot Singh

**Tester:** Harris Leung

**Editorialist:** Kanhaiya Mohan

# DIFFICULTY:

Cakewalk

# PREREQUISITES:

None

# PROBLEM:

There are 3 distinct regions in the city namely A, B, C consisting of P_A, P_B and P_C number of people respectively.

We know that people of the region B are in conflict with people of regions A and C. So, there will be a conflict if people from region B are present at the party along with people from region A or C.

You want to invite as many people as possible and also avoid any conflicts. Find the maximum number of people you can invite to the party.

# EXPLANATION:

We are given that the people of region B are in conflict with people of region A and C. This means that if people of region B come to the party, no one else comes.

On the other hand, if the people of region B do not come, people from regions A and C can come together.

The maximum number of people would be the maximum out of these two cases.

Thus, the answer is max(P_B, P_A +P_C).

# TIME COMPLEXITY:

The time complexity is O(1) per test case.

# SOLUTION:

## Tester's Solution

```
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
const ll mod=998244353;
int t;
int n,m;
ll a[300001];
ll s=0;
int main(){
ios::sync_with_stdio(false);cin.tie(0);
int t;cin >> t;
while(t--){
int x,y,z;
cin >> x >> y >> z;
cout << max(y,x+z) << '\n';
}
}
```

## Editorialist's Solution

```
#include <bits/stdc++.h>
using namespace std;
int main() {
int t;
cin>>t;
while(t--){
int pa, pb, pc;
cin>>pa>>pb>>pc;
cout<<max(pb, pa+pc)<<endl;
}
return 0;
}
```