Hduoj1358 【KMP】

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

Output

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.
Sample Input
3
aaa
12
aabaabaabaab
0
 

Sample Output
Test case #1
2 2
3 3

Test case #2
2 2
6 2
9 3
12 4

 

 

#include <stdio.h>
char s[1000005];
int next[1000005];
void getnext(int x) {
 int i=0,j=-1;
 next[0]=-1;
  while(i<x) {
   if(j==-1||s[i]==s[j]){
    i++;j++;
  //  if(s[i]==s[j])   这三行是next的优化，此处不需要
   //   next[i]=next[j];
  //  else
      next[i]=j;
   }
   else j=next[j];
  }
}

int main()
{
 int i,n,p=0;
 while(scanf("%d",&n),n) {
  scanf("%s",s);
  getnext(n);
  printf("Test case #%d\n",++p);
  for(i=2;i<=n;i++){      //关键在这里 
  //类似 len 如果能整除 len-next[len]，
  //那么重复的次数就是  len/(len-next[len])
  //不能整除就没有重复串
   int t=i%(i-next[i]);
   int k=i/(i-next[i]);
   if(t==0&&k>1) printf("%d %d\n",i,k);
  }
  puts("");
 }
}

//附上原理

kmp next函数 kmp的周期问题，深入了解kmp中next的原理

-----------------------

-----------------------

 k    m        x      j       i

由上，next【i】=j，两段红色的字符串相等（两个字符串完全相等），s[k....j]==s[m....i]

设s[x...j]=s[j....i](xj=ji)

则可得，以下简写字符串表达方式

kj=kx+xj;

mi=mj+ji;

因为xj=ji，所以kx=mj，如下图所示

 

-------------

      -------------

 k   m        x     j   

看到了没，此时又重复上面的模型了，kx=mj，所以可以一直这样递推下去

所以可以推出一个重要的性质len-next[i]为此字符串的最小循环节(i为字符串的结尾)，另外如果len%(len-next[i])==0,此字符串的最小周期就为len/(len-next[i]);

原文链接：http://blog.sina.com.cn/s/blog_792e6df50100wkps.html