Think about the case:
2,4,6 divisible by 2
3,6 divisible by 3
so unique numbers are 2,3,4 counted as 3
so that is win case
///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<string>
#include<cstring>
#include<sstream>
#include<cmath>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
#include<set>
#include<stack>
#include<vector>
#include<iterator>
#include <functional> ///sort(arr,arr+n,greater<int>()) for decrement of array
/*every external angle sum=360 degree
angle find using polygon hand(n) ((n-2)*180)/n*/
///Floor[Log(b) N] + 1 = the number of digits when any number is represented in base b
using namespace std;
typedef long long ll;
int main()
{
ll tst,a,b,k,n;
cin>>tst;
while(tst--)
{
cin>>n>>a>>b>>k;
ll lcm=(a*b)/__gcd(a,b);
if(((n/a)+(n/b)-(2*(n/lcm)))>=k) cout<<"Win"<<endl;
else cout<<"Lose"<<endl;
}
return 0;
}
///Alhamdulillah
problem link:https://www.codechef.com/LP2TO307/problems/HMAPPY2
Comments
Post a Comment