x is and Element of Array 1 and y is the element of array 2,↵
Find the number of ways in the two arrays where x^y>y^x (NOTE: ^ denotes power here and not the XOR)↵
↵
_First line inputs the number of test cases,↵
second line input the number of elements in two upcoming arrays↵
third line inputs the first array↵
fourth line inputs the second array_↵
↵
EXAMPLE:↵
sampleINPUT:↵
↵
~~~~~↵
1↵
3 2↵
2 1 6↵
1 5↵
~~~~~↵
↵
↵
↵
↵
sampleOUTPUT↵
3↵
↵
<<HERE IS MY POOR CODE,JUST A BRUTE FORCE!!>>↵
**IS THERE ANY WAY TO SOLVE IT IN TIME LESS THAN THIS??**↵
↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
↵
int main()↵
{↵
↵
int t;↵
cin>>t;↵
while(t--)↵
{↵
int m,n;↵
cin>>m>>n;↵
vector<int> vm(m,0);↵
vector<int> vn(n,0);↵
for(auto &it:vm)↵
cin>>it;↵
for(auto &it:vn)↵
cin>>it;↵
int counts=0;↵
for(int x:vm)↵
for(int y:vn)↵
{↵
↵
if(pow(x,y)>(pow(y,x)))↵
{↵
counts++;↵
↵
}↵
continue;↵
}↵
cout<<counts<<endl;↵
}↵
}↵
↵
~~~~~↵
↵
↵
Find the number of ways in the two arrays where x^y>y^x (NOTE: ^ denotes power here and not the XOR)↵
↵
_First line inputs the number of test cases,↵
second line input the number of elements in two upcoming arrays↵
third line inputs the first array↵
fourth line inputs the second array_↵
↵
EXAMPLE:↵
sampleINPUT:↵
↵
~~~~~↵
1↵
3 2↵
2 1 6↵
1 5↵
~~~~~↵
↵
↵
↵
↵
sampleOUTPUT↵
3↵
↵
<<HERE IS MY POOR CODE,JUST A BRUTE FORCE!!>>↵
**IS THERE ANY WAY TO SOLVE IT IN TIME LESS THAN THIS??**↵
↵
↵
~~~~~↵
#include<bits/stdc++.h>↵
using namespace std;↵
↵
int main()↵
{↵
↵
int t;↵
cin>>t;↵
while(t--)↵
{↵
int m,n;↵
cin>>m>>n;↵
vector<int> vm(m,0);↵
vector<int> vn(n,0);↵
for(auto &it:vm)↵
cin>>it;↵
for(auto &it:vn)↵
cin>>it;↵
int counts=0;↵
for(int x:vm)↵
for(int y:vn)↵
{↵
↵
if(pow(x,y)>(pow(y,x)))↵
{↵
counts++;↵
↵
}↵
continue;↵
}↵
cout<<counts<<endl;↵
}↵
}↵
↵
~~~~~↵
↵
↵