POJ-1002: 487-3279 详解2: 号码全集法

/*---------------------------------
 * 使用电话号码全集
 * --------------------------------*/
#include "stdio.h"
#include "string.h"
#include "stdlib.h"

/* 字母到数字的映射表 */
const int map[26] = {
    2, 2, 2,    /* A, B, C */
    3, 3, 3,    /* D, E, F */
    4, 4, 4,    /* G, H, I */
    5, 5, 5,    /* J, K, L */
    6, 6, 6,    /* M, N, O */
    7, 0, 7, 7, /* P, Q, R, S */
    8, 8, 8,    /* T, U, V */
    9, 9, 9,    /* W, X, Y */
    0           /* Z */
} ;

/* 电话号码全集 */
int teleNum[10000000] = {0} ;

int main(void)
{
    int  n = 0 ;
    char s[50] = {0} ;
    int  num = 0 ;
    int  i = 0 ;
    int  min = 10000000 ;
    int  max = -1 ;

    scanf("%d", &n) ;
    while(n--)
    {
        memset(s, 0, sizeof(s)) ;
        scanf("%s", s) ;        

        /* 将电话号码转换为整型 */
        num = 0 ;
        for(i = 0; i < sizeof(s); i++)
        {
            if(0 == s[i])
                break ;
            if(‘Q‘ == s[i] || ‘Z‘ == s[i])
                continue ;
            if(s[i] >= ‘A‘ && s[i] <= ‘Z‘)
                num = num * 10 + map[s[i] - ‘A‘] ;
            if(s[i] >= ‘0‘ && s[i] <= ‘9‘)
                num = num * 10 + s[i] - ‘0‘ ;
        }

        teleNum[num]++ ;
        if(2 == teleNum[num])
        {
            min = num < min ? num : min ;
            max = num > max ? num : max ;
        }
    }

    if(-1 == max)
    {
        printf("No duplicates.\r\n") ;
        return 0 ;
    }

    for(i = min; i <= max; i++)
    {
        if(teleNum[i] < 2)
            continue ;

        printf("%03d-%04d %d\r\n", i / 10000,
            i % 10000, teleNum[i]) ;
    }

    return 0 ;
}