poj1016 Numbers That Count

     這是一道簡單的模拟題，統計字元串中1,2....9的個數，同時組成一個新的字元串，并和最初的比較，有四種情況~~

    具體代碼如下：

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

char word[100],str[20][100],tempstr[100];
int count[10];
int index;

void change(int x)
{
    memset(count,0,sizeof(count));
    int len,i,j;
    len = strlen(str[x]);
    for(i = 0;i < len;i++)
       count[str[x][i]-'0']++;
    index = 0;
    for(i = 0;i < 10;i++){
        if(count[i]){
            if(count[i] > 9){
                tempstr[index++] = count[i]/10 + '0';
                tempstr[index++] = count[i]%10 + '0';
            }else{
                tempstr[index++] = count[i] + '0';
            }
            tempstr[index++] = i + '0';
        }
    }
    tempstr[index] = 0;
}

int main()
{
    int i,j;
    while(gets(str[0])){
        if(str[0][0]=='-') break;
        for(i = 1;i <= 15;i++){
            change(i-1);
        for(j = i-1;j>=0;j--){
            if(!memcmp(tempstr,str[j],strlen(str[j])>strlen(tempstr)?strlen(str[j]):strlen(tempstr)))
            {
                if(i == 1) {
                   printf("%s is self-inventorying\n",str[0]);
                }
                else if(j == i-1) {
                        printf("%s is self-inventorying after %d steps \n",str[0],i-1);
                        }
                else {
                    printf("%s enters an inventory loop of length %d\n",str[0],i - j);
                    }
                goto mark;
            }
        }
           strcpy(str[i],tempstr);
        }
        printf("%s can not be classified after 15 iterations\n",str[0]);
        mark:;
    }
    return 0;
}