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大致题意：
题意不难懂，对于任意的数字串n，都可以压缩存储为
c1 d1 c2 d2 .... ck dk 形式的数字串
而存在一些特别的数字串，其压缩前后的样子是一模一样的
定义这种数字串为self-inventorying
当我们把n看成原串，
A为n压缩1次后的数字串，
B为n压缩2次后的数字串（即A压缩1次后的数字串）
....以此类推
K为n压缩k次后的数字串（即K-1压缩k-1次后的数字串）
则可以延伸出数字串n的3种属性：
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
现给定一字符串，输出其属性。 属性1优于属性2，属性2优于属性3
解题思路：
字符串处理，纯粹的模拟题
压缩n时要注意，ck可能是1位，也可能是2位，需要判断。
设R(n)为描述整数n的压缩数字串
1、当R(n)==R(R(n))时，则n is self-inventorying
2、对于整数n：
令N = n
for j=1 to 15
{ 令tj =R(N)
若R(tj)== R(R(tj)) ，则n is self-inventorying after j steps 且 break
否则 N=tj
}
3、对于整数n：
令N = n，记录num[0]= n
for j=1 to 15
{ 令tj =R(N)，记录num[j]=tj
for i=0 to j-2 (保证k>=2)
{ 若tj== num[i] ，则n enters an inventory loop of length k ( k=j-i )
break
}
}
4、当且仅当n的3种属性都不存在时，n can not be classified after 15 iterations
Source修正
East Central North America 1998
http://plg1.cs.uwaterloo.ca/~acm00/regional98/real/
1 //Memory Time
2 //232K 32MS
3
4 #include<iostream>
5 #include<string>
6 using namespace std;
7
8 /*压缩数字串n，存放到t*/
9 void R(char* n,char* t)
10 {
11 int i,j;
12 int time[10]={0}; //记录n中各个数字出现的次数
13 for(i=0;n[i];i++)
14 time[ n[i]-'0' ]++;
15
16 for(i=0,j=0;i<10;i++)
17 {
18 if(time[i])
19 {
20 if(time[i]<10) //数字i出现次数<10，即占1位
21 {
22 t[j++]=time[i]+'0';
23 t[j++]=i+'0';
24 }
25 else //数字i出现次数>=10，即占2位
26 {
27 t[j++]=time[i]/10+'0';
28 t[j++]=time[i]%10+'0';
29 t[j++]=i+'0';
30 }
31 }
32 }
33 t[j]='\0';
34
35 return;
36 }
37
38 int main(int i,int j)
39 {
40 char n[16][85]; //n[0]为原串，n[1~15]分别为n连续压缩15次的数字串
41
42 while(cin>>n[0] && n[0][0]!='-')
43 {
44 bool flag1=false; //属性1，n is self-inventorying
45 int flag2=0; //属性2，n is self-inventorying after j steps，顺便记录j
46 int flag3=0; //属性3，n is enters an inventory loop of length k，顺便记录k
47
48 for(i=1;i<=15;i++)
49 R(n[i-1],n[i]);
50
51 if(!strcmp(n[0],n[1])) //属性1，n压缩1次就是其本身
52 flag1=true;
53
54 if(!flag1)
55 {
56 for(j=1;j<15;j++)
57 if(!strcmp(n[j],n[j+1])) //属性2, n压缩j次后的数字串n[j]具有属性1
58 {
59 flag2=j;
60 break;
61 }
62
63 if(!flag2)
64 {
65 for(j=1;j<=15;j++) //属性3，两两枚举各次压缩的数字串，注意循环间隔>=2
66 {
67 for(i=0;i<=j-2;i++)
68 {
69 if(!strcmp(n[j],n[i]))
70 {
71 flag3=j-i;
72 break;
73 }
74 }
75 if(flag3)
76 break;
77 }
78 }
79 }
80
81 if(flag1)
82 cout<<n[0]<<" is self-inventorying"<<endl;
83 else if(flag2)
84 cout<<n[0]<<" is self-inventorying after "<<flag2<<" steps"<<endl;
85 else if(flag3)
86 cout<<n[0]<<" enters an inventory loop of length "<<flag3<<endl;
87 else
88 cout<<n[0]<<" can not be classified after 15 iterations"<<endl;
89 }
90 return 0;
91 }
Sample Input
22
31123314
314213241519
21221314
111222234459
123456789
654641656
2101400052100005496
10000000002000000000
333
1
99999999999999999999999999999999999999999999999999999999999999999999999999999999
0000
0001
0111
1111
123456789
456137892
123213241561
543265544536464364
5412314454766464364
543267685643564364
5423434560121016464364
-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
123456789 is self-inventorying after 5 steps
654641656 enters an inventory loop of length 2
2101400052100005496 enters an inventory loop of length 2
10000000002000000000 enters an inventory loop of length 3
333 is self-inventorying after 11 steps
1 is self-inventorying after 12 steps
99999999999999999999999999999999999999999999999999999999999999999999999999999999 can not be classified after 15 iterations
0000 enters an inventory loop of length 2
0001 is self-inventorying after 8 steps
0111 is self-inventorying after 8 steps
1111 is self-inventorying after 8 steps
123456789 is self-inventorying after 5 steps
456137892 is self-inventorying after 5 steps
123213241561 enters an inventory loop of length 2
543265544536464364 enters an inventory loop of length 2
5412314454766464364 is self-inventorying after 3 steps
543267685643564364 enters an inventory loop of length 2
5423434560121016464364 is self-inventorying after 3 steps