假定一種編碼的編碼范圍是a-y的25個(gè)字母，從一位到四位的編碼，如果我們把該編碼按字典序排序，形成一個(gè)數(shù)組如下：
a，aa，aaa，aaaa，aaab，aaac，... ...，b，ba，baa，baaa，baab，baac，... ...，yyyw，yyyx，yyyy；
其中a的Index為0，aa的Index為1，aaa的Index為2，以此類推。
編寫一個(gè)函數(shù)，輸出給定編碼對(duì)應(yīng)Index。如：baca --> 16331.

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

#define N3 1
#define N2 25
#define N1 (25 * 25)
#define N0 (25 * 25 * 25)

#define C3 N3
#define C2 (N2 + N3)
#define C1 (N1 + N2 + N3)
#define C0 (N0 + N1 + N2 + N3)

int main()
{
    char code[11][5] = {"a", "aa", "aaa", "aaaa", "ab", "aba", "b", "ba", "bac", "baca", "yyyy"};
    
    for(int i = 0; i < 11; i++){
        int index = 0;
        switch(strlen(code[i])){
            case 1: index = C0 * (code[i][0] - 'a'); break;
            case 2: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + 1; break;
            case 3: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + 2; break;
            case 4: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + C3 * (code[i][3] - 'a') + 3; break;
            default: break;
        }
        printf("code = %-4s, strlen =%2d, index =%7d\n", code[i], strlen(code[i]), index);
    }
    return 0;   
}

結(jié)果顯示如下：

code = a   , strlen = 1, index =      0
code = aa  , strlen = 2, index =      1
code = aaa , strlen = 3, index =      2
code = aaaa, strlen = 4, index =      3
code = ab  , strlen = 2, index =    652
code = aba , strlen = 3, index =    653
code = b   , strlen = 1, index =  16276
code = ba  , strlen = 2, index =  16277
code = bac , strlen = 3, index =  16330
code = baca, strlen = 4, index =  16331
code = yyyy, strlen = 4, index = 406899

完整截圖：

編碼-騰訊機(jī)試題.png

改進(jìn)版代碼：

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

#define N3 1
#define N2 25
#define N1 (25 * 25)
#define N0 (25 * 25 * 25)

#define C3 N3
#define C2 (N2 + N3)
#define C1 (N1 + N2 + N3)
#define C0 (N0 + N1 + N2 + N3)

int main()
{
    char code[11][5] = {"a", "aa", "aaa", "aaaa", "ab", "aba", "b", "ba", "bac", "baca", "yyyy"};
    
    for(int i = 0; i < 11; i++){
        int index = 0;
        switch(strlen(code[i])){
            case 4: index += C3 * (code[i][3] - 'a') + 1;
            case 3: index += C2 * (code[i][2] - 'a') + 1;
            case 2: index += C1 * (code[i][1] - 'a') + 1;
            case 1: index += C0 * (code[i][0] - 'a');
        /*
            case 1: index = C0 * (code[i][0] - 'a'); break;
            case 2: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + 1; break;
            case 3: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + 2; break;
            case 4: index = C0 * (code[i][0] - 'a') + C1 * (code[i][1] - 'a') + C2 * (code[i][2] - 'a') + C3 * (code[i][3] - 'a') + 3; break;
        */
            default: break;
        }
        printf("code = %-4s, strlen =%2d, index =%7d\n", code[i], strlen(code[i]), index);
    }
    return 0;   
}

改進(jìn)版完整截圖：

改進(jìn)版完整截圖.png