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数字反转 题解:
P1276 上海交通大学机试题
发布于2026年3月19日 23:42
阅读数 30

#include<iostream>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
using namespace std;
int main() {
	int a, b;
	while (cin >> a >> b) {
		int arr[10000], brr[10000], index1 = 0, index2 = 0, sum1 = 0, sum2 = 0, sum = 0, index3 = 0, crr[10000];
		int num = a + b, N = a + b;
		while (a) {
			arr[index1++] = a % 10;
			a /= 10;
		}
		for (int i = 0;i < index1;i++) {
			sum1 = sum1 * 10 + arr[i];
		}
		while (b) {
			brr[index2++] = b % 10;
			b /= 10;
		}
		for (int i = 0;i < index2;i++) {
			sum2 = sum2 * 10 + brr[i];
		}
		while (num) {
			crr[index3++] = num % 10;
			num /= 10;
		}
		for (int i = 0;i < index3;i++) {
			sum = sum * 10 + crr[i];
		}
		if (sum1 + sum2 == sum) {
			cout << N << endl;
		}
		else {
			cout << "NO" << endl;
		}
	}
	return 0;
}

 

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