Maximum number of pairwise decrements possible in three numbers

Правка en1, от liveoverflow, 2020-03-29 00:21:51

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";
}
}

~~~~~

Теги #math, #algorithms

История

 
 
 
 
Правки
 
 
  Rev. Язык Кто Когда Δ Комментарий
en1 Английский liveoverflow 2020-03-29 00:21:51 1155 Initial revision (published)