Uvalive

参考：http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html

总结一下，如果对于next数组中的 i，符合 i % ( i - next[i] ) == 0 && next[i] != 0 , 则说明字符串循环，而且

循环节长度为: i - next[i]

循环次数为: i / ( i - next[i] )

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cmath>
 4 #include <vector>
 5 #include <cstring>
 6 #include <string>
 7 #include <algorithm>
 8 #include <string>
 9 #include <set>
10 #include <functional>
11 #include <numeric>
12 #include <sstream>
13 #include <stack>
14 #include <map>
15 #include <queue>
16 #pragma comment(linker, "/STACK:102400000,102400000")
17 #define CL(arr, val)    memset(arr, val, sizeof(arr))
18 
19 #define ll long long
20 #define inf 0x7f7f7f7f
21 #define lc l,m,rt<<1
22 #define rc m + 1,r,rt<<1|1
23 #define pi acos(-1.0)
24 
25 #define L(x)    (x) << 1
26 #define R(x)    (x) << 1 | 1
27 #define MID(l, r)   (l + r) >> 1
28 #define Min(x, y)   (x) < (y) ? (x) : (y)
29 #define Max(x, y)   (x) < (y) ? (y) : (x)
30 #define E(x)        (1 << (x))
31 #define iabs(x)     (x) < 0 ? -(x) : (x)
32 #define OUT(x)  printf("%I64d
", x)
33 #define lowbit(x)   (x)&(-x)
34 #define Read()  freopen("a.txt", "r", stdin)
35 #define Write() freopen("b.txt", "w", stdout);
36 #define maxn 1000000000
37 #define N 2510
38 #define mod 1000000000
39 using namespace std;
40 
41 char a[1000010];
42 int p[1000010];
43 void next(int l)
44 {
45     int j=0;
46     p[1]=0;
47     for(int i=2;i<=l;i++)
48     {
49         while(j>0&&(a[j+1]!=a[i])) j=p[j];
50         if(a[j+1]==a[i]) j+=1;
51         p[i]=j;
52     }
53 }
54 int main()
55 {
56     //freopen("a.txt","r",stdin);
57     int n,j=1;
58     while(~scanf("%d",&n))
59     {
60         if(n==0) break;
61         printf("Test case #%d
",j++);
62         scanf("%s",a+1);
63         int l=strlen(a+1);
64         next(l);
65         for(int i=1;i<=l;i++)
66         {
67             if(i%(i-p[i])==0&&p[i]!=0)
68             {
69                 printf("%d %d
",i,i/(i-p[i]));
70             }
71         }
72         printf("
");
73     }
74     return 0;
75 }