Lucky codeforces solution in c++

 A. Lucky?

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A ticket is a string consisting of six digits. A ticket is considered lucky if the sum of the first three digits is equal to the sum of the last three digits. Given a ticket, output if it is lucky or not. Note that a ticket can have leading zeroes.

Input

The first line of the input contains an integer t$t$ (1≤t≤103$1\le t\le {10}^3$) — the number of testcases.

The description of each test consists of one line containing one string consisting of six digits.

Output

Output t$t$ lines, each of which contains the answer to the corresponding test case. Output "YES" if the given ticket is lucky, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

Example

input

Copy

5
213132
973894
045207
000000
055776

output

Copy

YES
NO
YES
YES
NO

Note

In the first test case, the sum of the first three digits is 2+1+3=6$2+1+3=6$ and the sum of the last three digits is 1+3+2=6$1+3+2=6$, they are equal so the answer is "YES".

In the second test case, the sum of the first three digits is 9+7+3=19$9+7+3=19$ and the sum of the last three digits is 8+9+4=21$8+9+4=21$, they are not equal so the answer is "NO".

In the third test case, the sum of the first three digits is 0+4+5=9$0+4+5=9$ and the sum of the last three digits is 2+0+7=9$2+0+7=9$, they are equal so the answer is "YES".

solution<1>:

#include<iostream>
#include<climits>
#define ll long long
#include<vector>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
 
void solve()
{
	string s;
	cin >> s;
	ll cnt1 = 0, cnt2 = 0;
	ll n = s.length();
	for (ll i = 0; i < n/2; i++)
	{
		cnt1 += s[i] - '0';
		cnt2 += s[n - 1 - i] - '0';
	}
	(cnt1 == cnt2) ? cout << "YES\n" : cout << "NO\n";
 
}
 
int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		solve();
	}
	return 0;
}

solution<2>:

#include<iostream>
using namespace std;
 
void solve() {
	string s;
	cin >> s;
	if(s[0]+s[1]+s[2] == s[3]+s[4]+s[5]) {
		cout << "YES" << endl;
	}
	else {
		cout << "NO" << endl;
	}
}
int main() {
	int t ;
	cin >> t;
	while (t--) {
		solve();
	}
}