A XOR problem - HackerEarth Solution

A XOR problem - HackerEarth Solution

A XOR problem - HackerEarth Solution 

You are given two positive integers $X$  and $Y$  without leading zeroes. You can perform an operation on these integers any number of times, in which you can delete a digit of the given number such that resulting number does not have leading zeroes. Let $X^{\prime }$  and $Y^{\prime }$ be two numbers that were formed after performing operations on $X$  and $Y$ respectively.

You have to find all the unique pairs of $X^{\prime }$  and $Y^{\prime }$  whose XOR is zero.

Note: Two pairs of numbers $(A,B)$ and $(C,D)$ are considered different if and only if $A!=C$ or $B!=D$ .

Input format

• First line: An integer $T$  denoting the number of test cases
• Each of the next $T$  lines: Two space-separated integers $X$  and $Y$ 

Output format

• For each test case, print the answer in a new line.

Constraints

$$\begin{gathered} 1\le T\le 10 \\ 1\le X,Y\le {10}^{12} \end{gathered}$$

Sample Input

2
12 102
13 33

Sample Output

3
1

Sample Explanation

 For the test case 1,

• Possible values of X' are [1,2,12]
• Possible values of Y' are [1,2,10,12,102]

So, three pairs (1,1), (2,2), (12,12) have xor zero.

For the test case 2,

• Possible values of X' are [1,3,13]
• Possible values of Y' are [3,33] (We can generate 3 two times)

So, one unique pair (3,3) has xor zero.

A XOR problem - HackerEarth Solution  

Code:

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

typedef long long int LL;

int main() 
{
	//freopen("in1.txt","r",stdin);
	//freopen("out1.txt","w",stdout);
	int t;
	cin>>t;
	while(t--){
	LL x,y;
	cin>>x>>y;
	string X,Y;
	X = to_string(x);
	Y = to_string(y);
	unordered_map<LL,LL> um;
	for(int i=0;i<(1<<X.length());i++)
	{
		LL num = 0;
		for(int j=0;j<X.length();j++)
		{
			if(i&(1<<j))
			{
				num = num*10 + (X[j]-'0');
			}
		}
		if(num!=0)
		{
			um[num]++;
		}
	}
	set<LL> st;
	for(int i=0;i<(1<<Y.length());i++)
	{
		LL num = 0;
		for(int j=0;j<Y.length();j++)
		{
			if(i&(1<<j))
			{
				num = num*10 + (Y[j]-'0');
			}
		}
		st.insert(num);
	}
	LL ans=0;
	for(auto k:st)
	{
		if(um[k])
		{
			ans++;
		}
	}
	cout<<ans<<"\n";}
	return 0;
}

A XOR problem - HackerEarth Solution 

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