hdu 1358 Period

Description

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) �C 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

 3aaa12aabaabaabaab0 

 

Sample Output

 Test case #12 23 3Test case #22 26 29 312 4 

 

解释一下题意：告诉你一个字符串，然后

Test case #22 26 29 312 4

第二行的2 2，意思是，从第1个字母到第2字母组成的字符串可由某一周期性的字串（“a”）

的两次组成，也就是aa有两个a组成；

第三行自然就是aabaab可有两个aab组成；

第四行aabaabaab可由三个aab组成；

第五行aabaabaabaab可有四个aab组成

对next数组还不是很理解的可以去next 数组深入理解看一下

另外说一下next[0]只是一个边界，防止越界和判断用，实际next[1]对应的才是数组的第一个。

#include<stdio.h>#include<string.h>using namespace std;int next[1000002];char b[1000002];int n;void getn(){    int i,j;    i=0;    j=-1;    next[0]=-1;    int len=strlen(b);    while(i<len){        if(j==-1||b[i]==b[j])        {            i++;            j++;            next[i]=j;        }        else            j=next[j];     }}int main(){    int k=1;   while(scanf("%d",&n)){        if(n==0)        break;        memset(next,0,sizeof(next));        memset(b,'0',sizeof(b));      scanf("%s",b);         getn();    printf("Test case #%d\n",k++);    int i;    for(i=2;i<=n;i++){        if(next[i]!=0&&i%(i-next[i])==0)            printf("%d %d\n",i,i/(i-next[i]));    }      printf("\n");   }   return 0;}