题解 | #特殊乘法#

#include<bits/stdc++.h>
using namespace std;
int main() {
    long int shu1 = 0, shu2 = 0;
    while (cin >> shu1 >> shu2) {
        int arr1[10] = {}, arr2[10] = {};
        int i = 0, j = 0;
        int sum = 0;
        while (shu1 != 0) {
            arr1[i] = shu1 % 10;
            shu1 = shu1 / 10;
            i++;
        }
        while (shu2 != 0) {
            arr2[j] = shu2 % 10;
            shu2 = shu2 / 10;
            j++;
        }
        for (int k = 0; k < sizeof(arr1) / sizeof(arr1[0]); k++) {
            for (int l = 0; l < sizeof(arr2) / sizeof(arr2[0]);
                    l++) { //如何快速得到数组中元素个数
                sum = sum + arr1[k] * arr2[l];
            }
        }
        cout << sum << endl;
    }
    return 0;
}