Code is given below

#1

#include<bits/stdc++.h>
using namespace std;
vector str(string str)
{
vectorv;
int n;
n=str.length();
for(int i=0;i<n;i++)
v.push_back(str[i]);
return v;

}
int main()
{
int m;
cin>>m;
char c1[100];
char c2[100];
int n;
vectorv;
vectorv1;
vectorv2;
vector<pair<vector,vector>>v3;
vector<vector>v4;
for(int k=0;k<m;k++)
{
cin>>c1>>c2;
v1=str(c1);
v2=str(c2);
v3.push_back(make_pair(v1,v2));
}
n=v1.size();
for(int x=0;x<m;x++)
{
for(int i=0;i<n;i++)
v.push_back((v3[x].first)[i]^(v3[x].second)[i]);
v4.push_back(v);
v.clear();
}
for(auto x:v4)
{
for(int l=0;l<n;l++)
cout<<x[l];
cout<<endl;
}
return 0;
}

:bulb: Ultra-Fast-Mathematicians

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#2

hey @CHANDU_0556 you dont need to make your code this complex the approach is pretty simple if s1[i] is equal to s2[i] print 0, else 1.

#include<bits/stdc++.h>
using namespace std;
int main() {
	int t;
	cin >> t;
	while(t--){
		string s1, s2;
		cin >> s1 >> s2;
		for(int i = 0; i < s1.length(); i++){
			if(s1[i] == s2[i])
				cout << '0';
			else
				cout << '1';
		}
		cout << endl;
	}
}
#3

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