题解 | #字符串加解密#

思路

1. 逐串分析，暴力破解
2. 根据题目描述，逐步实现加密解密

Answer

#include<stdio.h>
#include<string.h>

int main() {
    char str[1000];
    char str_un[1000];
    scanf("%s", str);
    scanf("%s", str_un);
    int len = strlen(str);
    int len_un = strlen(str_un);

    for (int i = 0; i < len; i++) {
        if (str[i] >= 'a' && str[i] <= 'z') {
            if (str[i] == 'z') {
                str[i] = 'A';
            } else{
                 str[i] = str[i]-32 + 1;
            }
               
        }else if (str[i] >= 'A' && str[i] <= 'Z') {
            if (str[i] == 'Z') {
                str[i] = 'a';
            } else{
                str[i] = str[i]+32 + 1;
            }
                

        }else if (str[i] >= '0' && str[i] <= '9') {
            if (str[i] == '9') {
                str[i] = '0';
            } else{
                 str[i] += 1;
            }
               
        }
        printf("%c", str[i]);
    }
    printf("\n");
    for (int i = 0; i < len_un; i++) {
        if (str_un[i] >= 'a' && str_un[i] <= 'z') {
            if (str_un[i] == 'a') {
                str_un[i] = 'Z';
            } else{
                str_un[i] = str_un[i]-32 - 1;
            }
                
        }else if (str_un[i] >= 'A' && str_un[i] <= 'Z') {
            if (str_un[i] == 'A') {
                str_un[i] = 'z';
            } else{
                str_un[i] = str_un[i] +32 - 1;
            }
                

        }else if (str_un[i] >= '0' && str_un[i] <= '9') {
            if (str_un[i] == '0') {
                str_un[i] = '9';
            } else{
                 str_un[i] -= 1;
            }
               
        }
        printf("%c", str_un[i]);


    }
    printf("\n");


    return 0;
}