 # MEENA_009 Editorial

Practice

Author: Ram Agrawal
Tester: Yogesh Deolalkar
Editorialist:Prathamesh Sogale

# DIFFICULTY:

CAKEWALK, SIMPLE.

Greedy, Math

# PROBLEM:

Meena has given two integers A,B. You have to choose an integer X in the range [minimum(A,B), maximum(A,B)] such that:

⌈B−X⌉/2 +⌈X−A⌉ /2 is maximum.

What is the maximum sum Meena can obtain if you choose X optimally?

Note: ⌈x⌉ : Returns the smallest integer that is greater than or equal to x (i.e rounds up to the nearest integer). For example, ⌈1.4⌉=2, ⌈5⌉=5, ⌈−1.5⌉=−1, ⌈−3⌉=−3 , ⌈0⌉=0.

``````
# SOLUTIONS:

[details="Setter's Solution"]
import math
for t in range(int(input())):
a,b=map(int,input().split())
max1=max(a,b)
min1=min(a,b)
mx=-100000

for i in range(min1,max1+1):
ans=math.ceil(math.ceil((b-i)/2)+math.ceil((i-a)/2))
mx=max(mx,ans)
print(mx)

[/details]

[details="Tester's Solution"]
#include <iostream>
using namespace std;

int main() {
int k;
cin>>k;
while(k--){
long long int a,b;
cin>>a>>b;
long long int diff=b-a;
if(a==b){
cout<<"0"<<endl;
}
else if(b>a){
cout<<(diff/2)+1<<endl;
}
else{
if(diff%2==0){
cout<<(diff/2)+1<<endl;
}
else{
cout<<(diff/2)<<endl;
}
}
}
return 0;
}

[/details]``````