一道耗时间的水题

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Problem Description:

    Do you know how to read the phone numbers in English? Now let me tell you.


For example, In China, the phone numbers are 11 digits, like: 15012233444. Someone divides the numbers into 3-4-4 format, i.e. 150 1223 3444. While someone divides the numbers into 3-3-5 format, i.e. 150 122 33444. Different formats lead to different ways to read these numbers:


150 1223 3444 reads one five zero one double two three three triple four.


150 122 33444 reads one five zero one double two double three triple four.


Here comes the problem:


Given a list of phone numbers and the dividing formats, output the right ways to read these numbers.


Rules:


Single numbers just read them separately.


2 successive numbers use double.


3 successive numbers use triple.


4 successive numbers use quadruple.


5 successive numbers use quintuple.


6 successive numbers use sextuple.


7 successive numbers use septuple.


8 successive numbers use octuple.


9 successive numbers use nonuple.


10 successive numbers use decuple.

More than 10 successive numbers read them all separately.

Input:


The first line of the input gives the number of test cases, T(1 ≤ T ≤ 100).


T test cases follow. Each line contains a phone number N(1 ≤ length of N ≤ 100) and the dividing format F, one or more positive integers separated by dashes (-), without zeros and whose sum always equals the number of digits in the phone number.


Output:


For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the reading sentence in English whose words are separated by a space.
//此题AC率没有看起来那么高，截至2015-04-05，仅有三人AC
Sample Input:


3
15012233444 3-4-4
15012233444 3-3-5
12223 2-3
Sample Output:


Case #1: one five zero one double two three three triple four
Case #2: one five zero one double two double three triple four
Case #3: one two double two three

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

#define INF 0x3f3f3f3f

char s[205];
char divi[505];
int a[200];
char rule[20][50]={"","","double","triple","quadruple","quintuple","sextuple","septuple","octuple","nonuple","decuple"};
char digit[15][20]={"zero","one","two","three","four","five","six","seven","eight","nine"};

void solve(int p,int q)
{
    int i,j,k=0;
    for(i=p;i<q;i++)
    {
        j=1;
        while(i+j<q&&s[i+j]==s[i])
        {
            j++;
        }
        if(j<2)
            printf(" %s",digit[s[i]-'0']);
        else if(j<=10)
            printf(" %s %s",rule[j],digit[s[i]-'0']);
        else
        {
            for(k=0;k<j;k++)
                printf(" %s",digit[s[i]-'0']);
        }
        i+=j-1;
    }
}

int main()
{
    int i,j,k,n,len,num;
    int t,ca=1;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",s,divi);
        len=strlen(divi);
        memset(a,0,sizeof(a));
        for(i=0,num=0;i<len;i++)
        {
            while(divi[i]!='-'&&i<len)
            {
                a[num]=10*a[num]+divi[i]-'0';
                i++;
            }
            num++;
        }
        /*for(i=0;i<num;i++)
           printf("%d ",a[i]);*/
        //len=strlen(s);
        printf("Case #%d:",ca++);
        for(i=0,k=0;i<num;i++)
        {
            solve(k,k+a[i]);
            k+=a[i];
        }
        printf("\n");
    }
    return 0;
}