### liveoverflow's blog

By liveoverflow, history, 2 months ago, ,

Given are the 3 non-negative integers a,b,c

In a single operation, we have to subtract 1 from two integers only if they don't become negative.

We have to find the maximum no of operations possible that we can do until the given operation is not possible.

constraints:1<=a,b,c<=10^18 , 1<=test-cases<=10^5

Example- (1,2,3) = (1,1,2) -> (1,0,1) -> (0,0,0) , ans is 3

(1,1,8) = (1,0,7) -> (0,0,6) , ans is 2

Any approach or proof will be highly helpful.

I have actually written a code that's working as far as I know, but I don't know if it's completely true,

Thanks ~~~~~

#include<bits/stdc++.h>
using namespace std;

#define fastio ios_base::sync_with_stdio(0); cin.tie(0)
#define LL long long

int main(){
fastio;
int t; cin>>t;
while(t--){
LL a[3]; cin>>a[0]>>a[1]>>a[2];
sort(a,a+3);
if(a[0]+a[1]>=a[2]){
LL ans = a[2] + (a[0]+a[1]-a[2])/2;
cout<<ans;
}
else {
LL ans = a[1] + min(a[0],a[2]-a[1]);
cout<<ans;
}
cout<<"\n";
}
}

~~~~~

• 0

 » 2 months ago, # |   0 Let $(a, b, c)$ represent the input sorted in non-decreasing order (default std::sort). This means that $c$ must be greater than (or maybe equal to) $b$ and $b$ must be greater than (or maybe equal to) $a$. So, if $a + b <= c$, answer is simply $a + b$. Otherwise, when $a + b > c$, answer is $\frac{(a + b + c)}{2}$ (integer division).
 » 2 months ago, # |   0 If you want somewhere (else) to submit:https://codeforces.com/contest/1263/problem/A