POJ 1016 Numbers That Count（水~）

Description
每个数字串都可以被压缩成c1 d1 c2 d2 … cn dn的形式，如果一个串被压缩几次之后和自己相同则称之为self-inventorying，对于一个串n有三种情况：
1. n压缩1次就马上出现self-inventorying特性，即 n n n n n n n …..
2. n压缩j次后的数字串J出现self-inventorying特性，即 n A B C….H I J J J J J J J
3. n压缩j次后的数字串J，每再压缩K次，重新出现数字串J，即n A B… J ..K J ..K J..K J
其中K称为循环间隔，K>=2
现给出一个数字串，判断其属于以上哪一种
Input
多组用例，每组用例占一行为一长度不超过80的数字串，以-1结束输入
Output
对于每组用例，判断该串属于哪种情况
Sample Input
22
31123314
314213241519
21221314
111222234459
-1
Sample Output
22 is self-inventorying
31123314 is self-inventorying
314213241519 enters an inventory loop of length 2
21221314 is self-inventorying after 2 steps
111222234459 enters an inventory loop of length 2
Solution
水题
Code

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;#define maxn 111char s[22][maxn];void deal(char *s1,char *s2){    int len=strlen(s1),num[11],res=0;    memset(num,0,sizeof(num));    for(int i=0;i<len;i++)        num[s1[i]-'0']++;    for(int i=0;i<10;i++)        if(num[i])        {            if(num[i]>9)s2[res++]=num[i]/10+'0',num[i]%=10;            s2[res++]=num[i]+'0';            s2[res++]=i+'0';        }    s2[res]='\0';}int main(){    while(~scanf("%s",s[0]),s[0][0]!='-')    {        int f1,f2,f3;        f1=f2=f3=0;        for(int i=1;i<=15;i++)deal(s[i-1],s[i]);        if(!strcmp(s[0],s[1]))f1=1;        if(!f1)        {            for(int i=1;i<15;i++)                if(!strcmp(s[i],s[i+1]))                {                    f2=i;                    break;                }            if(!f2)            {                for(int i=1;i<=15;i++)                {                    for(int j=0;j<=i-2;j++)                        if(!strcmp(s[i],s[j]))                        {                            f3=i-j;                            break;                        }                    if(f3)break;                }            }        }        if(f1)printf("%s is self-inventorying\n",s[0]);        else if(f2)printf("%s is self-inventorying after %d steps\n",s[0],f2);        else if(f3)printf("%s enters an inventory loop of length %d\n",s[0],f3);        else printf("%s can not be classified after 15 iterations\n",s[0]);    }    return 0;}