题解 | 数字反转

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

int main() {
    int a, b;
    while (cin >> a >> b) { // 注意 while 处理多个 case
        int a1 = 0, b1 = 0, sum = a + b;
        while (a) {
            a1 = a1 * 10 + a % 10;
            a /= 10;
        }
        while (b) {
            b1 = b1 * 10 + b % 10;
            b /= 10;
        }
        a1 += b1;
        b1 = 0;
        while (a1) {
            b1 = b1 * 10 + a1 % 10;
            a1 /= 10;
        }
        if (b1 == sum) {
            cout << sum << endl;
        } else {
            cout << "NO" << endl;
        }
    }
}
// 64 位输出请用 printf("%lld")