AGOC - Editorial

editorial
,
1

PROBLEM LINK:

Contest

Setter: a0n0k0i0t
Tester: a0n0k0i0t
Editorialist: a0n0k0i0t

DIFFICULTY:

Easy

PREREQUISITES:

Basic Maths

Explanation:

Observation:

The most optimum approach to reach 0 is obivously to use 5 to subtract maximum possible times.
For a number N let X = N % 5
So,
if (X is even):
you can subtract 5 unti N is X
if(X in odd):
you can subtract 5 until N is X+5

rest of the operations will be subtracting 2.

Time complexity:

O(1) for each test case

SOLUTION

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

int main(){
    ll T;
    cin>>T;
    for(int _=0;_<T;_++){
        ll x,y;
        cin>>x>>y;
        ll cnt1=0,cnt2=0;
        ll a=x/5;
        cnt1+=a;
        x-=a*5;
        if(x%2){
            x+=5;
            cnt1--;
        }
        cnt1+=x/2;
        
        ll b=y/5;
        cnt2+=b;
        y-=b*5;
        if(y%2){
            y+=5;
            cnt2--;
        }
        cnt2+=y/2;
        
        if(cnt1<cnt2)cout<<"Ram\n";
        else if(cnt1>cnt2)cout<<"Shyam\n";
        else cout<<"Draw\n";
        
    }
    return 0;
}
2

What’s up it’s me, I am also visiting this website daily, this website is really pleasant and the viewers are really sharing good thoughts.