POJ 2608 Soundex（我的水题之路—

POJ 2608 Soundex（我的水题之路——字符值）

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发布时间：2019-05-25

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Soundex

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8200		Accepted: 4039

Description

Soundex coding groups together words that appear to sound alike based on their spelling. For example, "can" and "khawn", "con" and "gone" would be equivalent under Soundex coding.

Soundex coding involves translating each word into a series of digits in which each digit represents a letter:

1 represents B, F, P, or V      2 represents C, G, J, K, Q, S, X,  or Z      3 represents D or T      4 represents L      5 represents M or N      6 represents R

The letters A, E, I, O, U, H, W, and Y are not represented in Soundex coding, and repeated letters with the same code digit are represented by a single instance of that digit. Words with the same Soundex coding are considered equivalent.

Input

Each line of input contains a single word, all upper case, less than 20 letters long.

Output

For each line of input, produce a line of output giving the Soundex code.

Sample Input

KHAWNPFISTERBOBBY

Sample Output

25123611

Source

26个字母分别对应了不同的值，现在给一个字符串，按照如下规则输出：

1）如果字母有值，则考虑输出;否则不输出。

2）多个拥有相同值的字母并列，仅输出一个值。

直接比较，如果对每一个字符进行取值，取上一次不同字符值的值last，进行比较，如果本字符有值（不为0），且不同于上一次的值last，就输出其值。

注意点：

1）注意打表正确（1WA）。

2）无论这个值是否输出，皆保存本次值为last。

代码（1AC 1WA）：

#include 
   
    #include 
    
     #include 
     
      char str[30];int arr[50];void init(){    arr[0] = 0;  // A    arr[1] = 1;  // B    arr[2] = 2;  // C    arr[3] = 3;  // D    arr[4] = 0;  // E    arr[5] = 1;  // F    arr[6] = 2;  // G    arr[7] = 0;  // H    arr[8] = 0;  // I    arr[9] = 2;  // J    arr[10] = 2; // K    arr[11] = 4; // L    arr[12] = 5; // M    arr[13] = 5; // N    arr[14] = 0; // O    arr[15] = 1; // P    arr[16] = 2; // Q    arr[17] = 6; // R    arr[18] = 2; // S    arr[19] = 3; // T    arr[20] = 0; // U    arr[21] = 1; // V    arr[22] = 0; // W    arr[23] = 2; // X    arr[24] = 0; // Y    arr[25] = 2; // Z}int main(void){    char last, tmp;    int len, i;    init();    while (scanf("%s", str) != EOF){        len = strlen(str);        last = 0;        for (i = 0; i < len ; i++){            if (arr[str[i] - 'A'] != 0 && arr[str[i] - 'A'] != last){                printf("%d", (last = arr[str[i] - 'A']));            }            else{                last = arr[str[i] - 'A'];            }        }        printf("\n");    }    return 0;}

注意点：

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