PAT 1048数字加密

#include<iostream>
#include<cstring>
using namespace std;
const int MAXN = 102;
char A[MAXN], B[MAXN], result[MAXN] = {0};

void reverse(char c[]) {
    int len = strlen(c);
    for (int i = 0; i < len / 2; i++) {
        char temp = c[i];
        c[i] = c[len - 1 - i];
        c[len - 1 - i] = temp;
    }
}

int main() {
    cin >> A >> B;
    //将A和B反转,要从低位开始
    reverse(A);
    reverse(B);
    int maxlen = strlen(A) > strlen(B) ? strlen(A) : strlen(B);
    for (int i = 0; i < maxlen; i++) {
        //计算当前A,B位的数字，有某一个的字符长度比较短，填0
        int numA = i < strlen(A) ? A[i] - ‘0‘ : 0;
        int numB = i < strlen(B) ? B[i] - ‘0‘ : 0;
        //偶数位
        if ((i + 1) % 2 == 0) {
            int c = numB - numA;
            if (c < 0) c += 10;
            result[i] = c + ‘0‘;
        }
        else {
            //奇数位
            int temp = (numA + numB) % 13;
            if (temp < 10) {
                result[i] = temp + ‘0‘;
            }
            else if(temp == 10){
                result[i] = ‘J‘;
            }
            else if (temp == 11) {
                result[i] = ‘Q‘;
            }
            else if (temp == 12) {
                result[i] = ‘K‘;
            }
        }
    }
    //将结果反转
    reverse(result);
    printf("%s\n", result);

    system("pause");
    return 0;
}