题解 | #数字反转#简单判断一下就行

//C++版代码
#include <iostream>
using namespace std;
int reverseNum(int x) {
    int ret = 0;
    while (x) {
        ret = ret * 10 + x % 10;
        x /= 10;
    }
    return ret;
}
int main() {
    int a, b;
    while (cin >> a >> b) {
        if (reverseNum(a) + reverseNum(b) == reverseNum(a + b)) cout << a + b << endl;
        else cout << "NO" << endl;
    }
    return 0;
}
//Java版代码
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNextInt()) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            if (reverseNum(a) + reverseNum(b) == reverseNum(a + b)) System.out.println(a + b);
            else System.out.println("NO");
        }
    }
    private static int reverseNum(int x) {
        return Integer.parseInt(new StringBuilder(Integer.toString(x)).reverse().toString());
    }
}
#Python版代码
while True:
    try:
        a, b = input().split()
        print(int(a) + int(b) if int(a[::-1]) + int(b[::-1]) == int(str(int(a) + int(b))[::-1]) else 'NO')
    except:
        break