Period HDU

   

PeriodHDU - 1358

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

3aaa12aabaabaabaab0

Sample Output

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

思路：这道题给定一个字符串，判断这个字符串任一前缀是不是周期的字符串，如果是输出这个字符串的长度和周期并且长度从小到大输出，通过KMP算法求出next数组这个题就解决了

code：

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int Next[1010000];char a[1010000];void getNext(){    int i = -1,j = 0;//两个下标错开    Next[0] = -1;//第一个初始化为-1    int lena = strlen(a);//求出字符串长度    while(j < lena){        if(i==-1||a[i]==a[j]){            i++,j++;            //if(a[i]==a[j])Next[j] = Next[i];//优化            Next[j] = i;//注意注意！！因为这里是单纯的找前缀中最长的前后缀相同长度，所以不能进行优化，优化长度就改变了        }        else i = Next[i];    }}int main(){    int n;    int i,j;    int num = 0;    while(~scanf("%d",&n)&&n){        scanf("%s",a);        getNext();        printf("Test case #%d\n",++num);        for(i = 1; i <= n; i++){            int temp = i-Next[i];//这句话求循环节的长度，仔细想想你会发现，如果不循环，那么i-Next[i]的值肯定大于前缀总长度的一般，而如果循环肯定小于等于一半            if(i%temp==0&&i/temp>1){//能整除说明是周期字符串，并且题目要求周期大于1                printf("%d %d\n",i,i/temp);            }        }        printf("\n");    }    return 0;}