Uvalive - 3026 Period (kmp求字符串的最小循环节+最大重复次数)

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] )

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 2510
 #define mod 1000000000
 using namespace std;

 char a[];
 int p[];
 void next(int l)
 {
     int j=;
     p[]=;
     for(int i=;i<=l;i++)
     {
         while(j>&&(a[j+]!=a[i])) j=p[j];
         if(a[j+]==a[i]) j+=;
         p[i]=j;
     }
 }
 int main()
 {
     //freopen("a.txt","r",stdin);
     int n,j=;
     while(~scanf("%d",&n))
     {
         if(n==) break;
         printf("Test case #%d\n",j++);
         scanf("%s",a+);
         int l=strlen(a+);
         next(l);
         for(int i=;i<=l;i++)
         {
             if(i%(i-p[i])==&&p[i]!=)
             {
                 printf("%d %d\n",i,i/(i-p[i]));
             }
         }
         printf("\n");
     }
     return ;
 }