CodeForces - 1676A A - Lucky?

最新推荐文章于 2025-05-29 23:36:29 发布

Lyriv_

最新推荐文章于 2025-05-29 23:36:29 发布

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本文链接：https://blog.youkuaiyun.com/qq_60775360/article/details/125960215

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该博客主要涉及一个简单的算法问题，即检查一个六位数的彩票号码是否幸运。一个幸运号码的定义是前三个数字之和等于后三个数字之和。通过输入测试用例，程序会判断每个号码是否满足这一条件，并输出'YES'或'NO'。示例解释了如何根据给定条件检查特定的彩票号码是否幸运。

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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 tt (1 \leq t \leq 10^31≤t≤10
3
) — the number of testcases.

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

Output
Output tt 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).

Sample 1
Inputcopy Outputcopy
5
213132
973894
045207
000000
055776
YES
NO
YES
YES
NO
Note
In the first test case, the sum of the first three digits is 2 + 1 + 3 = 62+1+3=6 and the sum of the last three digits is 1 + 3 + 2 = 61+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 = 199+7+3=19 and the sum of the last three digits is 8 + 9 + 4 = 218+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 = 90+4+5=9 and the sum of the last three digits is 2 + 0 + 7 = 92+0+7=9, they are equal so the answer is “YES”.

#include<iostream>
using namespace std;
int main(void) {
	int n;
	string a;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a;
		if (a[0] + a[1] + a[2] == a[3] + a[4] + a[5]) {
			cout << "YES" << endl;
		}
		else {
			cout << "NO" << endl;
		}
	}
}