题解 | 字符串的反码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>
#include <map>
using namespace std;
//'a' 与‘z' 差25  ’b'与

map<char, int> Rword = {
    {'a', 25}, {'b', 23}, {'c', 21}, {'d', 19}, {'e', 17},
    {'f', 15}, {'g', 13}, {'h', 11}, {'i', 9}, {'j', 7},
    {'k', 5}, {'l', 3}, {'m', 1}, {'n', -1}, {'o', -3},
    {'p', -5}, {'q', -7}, {'r', -9}, {'s', -11}, {'t', -13},
    {'u', -15}, {'v', -17}, {'w', -19}, {'x', -21}, {'y', -23},
    {'z', -25},
};
map<char, int> Rword02 = {
    {'A', 25}, {'B', 23}, {'C', 21}, {'D', 19}, {'E', 17},
    {'F', 15}, {'G', 13}, {'H', 11}, {'I', 9}, {'J', 7},
    {'K', 5}, {'L', 3}, {'M', 1}, {'N', -1}, {'O', -3},
    {'P', -5}, {'Q', -7}, {'R', -9}, {'S', -11}, {'T', -13},
    {'U', -15}, {'V', -17}, {'W', -19}, {'X', -21}, {'Y', -23},
    {'Z', -25},
};

int main() {
    char str1[81];
    scanf("%s", str1);
    for (int i = 0; str1[i] != '\0'; i++) {
        if (str1[i] >= 'a' && str1[i] <= 'z') {
            str1[i] += Rword[str1[i]];
        } else if (str1[i] >= 'A' && str1[i] <= 'Z') {
            str1[i] += Rword02[str1[i]];
        }
    }
    printf("%s\n", str1);

    return 0;
}