POJ -2406 power strings --KMP中next函数的应用

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 23370		Accepted: 9811

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

  1 /* 功能Function Description:     POJ-2406
  2    开发环境Environment:          DEV C++ 4.9.9.1
  3    技术特点Technique:
  4    版本Version:
  5    作者Author:                   可笑痴狂
  6    日期Date:                      20120813
  7    备注Notes:
  8    题意：
  9         找出字符串的最小循环周期从而算出最大循环次数
 10 */
 11 /*
 12 //代码一：-----简单枚举法
 13 #include<stdio.h>
 14 #include<string.h>
 15 char s[1000005];
 16 
 17 int main()
 18 {
 19     int i,len,k,j,flag;
 20     while(scanf("%s",s+1)&&s[1]!='.')
 21     {
 22         len=strlen(s+1);
 23     //    printf("%d\n%c",len,s[1]);
 24         for(i=1;i<=len;++i)
 25         {
 26             if(len%i)
 27                 continue;
 28             flag=1;
 29             for(k=1;flag&&k<=i;++k)
 30             {
 31                 for(j=2;j<=len/i;++j)
 32                 {
 33                     if(s[k]!=s[i*(j-1)+k])
 34                     {
 35                         flag=0;
 36                         break;
 37                     }
 38                 }
 39             }
 40             if(flag)
 41             {
 42                 printf("%d\n",len/i);
 43                 break;
 44             }
 45         }
 46     }
 47     return 0;
 48 }
 49 */
 50 
 51 //代码二：-------KMP算法中next函数的性质
 52 /*
 53     设Len是s的长度
 54     给出结论：
 55     如果len%(len-next[len])==0,则字符串中必存在最小循环节，且循环次数即为 len/(len-next[len])
 56     
 57     题目大意：给出一个字符串，求出是某个字符串的n次方
 58     解决：KMP找出字符串的循环节
 59     若有字符串ai...a(j+m)，若next[len+1]=m（字符串从1开始）,意味着从ai ...am 和 aj...a(j+m)相等，
 60     假设中间有相交部分（或者相交部分为0）aj....am，若总长度减去相交部分剩余的部分能被总长度整除，
 61     又因为  从ai ...am 和 aj...aj+m相等，所以除去相交部分剩余的的两段字符串相等，而且必定满足中间相交部分也能被起始字符串整除，依照这样，
 62     即可得出最小的循环节 
 63 
 64 */
 65 
 66 #include<stdio.h>
 67 #include<string.h>
 68 char s[1000005];
 69 int next[1000005];
 70 
 71 void get_next(int len)
 72 {
 73     int i,j;
 74     i=0;
 75     j=-1;
 76     next[0]=-1;
 77     while(i<len)
 78     {
 79         if(j==-1||s[i]==s[j])
 80         {
 81             ++i;
 82             ++j;
 83             next[i]=j;
 84         }
 85         else
 86             j=next[j];
 87     }
 88 
 89     /*
 90     while(i<len)              //改进算法---改进在某些情况下的缺陷  详见数据结构课本P84
 91     {
 92         if(j==-1||s[i]==s[j])
 93         {
 94             ++i;
 95             ++j;
 96             if(s[i]!=s[j])
 97                 next[i]=j;
 98             else
 99                 next[i]=next[j];
100         }
101         else
102             j=next[j];
103     }
104     */
105 }
106 
107 int main()
108 {
109     int i,len,k,j,flag;
110     while(scanf("%s",s)&&s[0]!='.')
111     {
112         len=strlen(s);
113         get_next(len);
114         if(len%(len-next[len])==0)
115             printf("%d\n",len/(len-next[len]));
116         else
117             printf("1\n");
118     }
119     return 0;
120 }