Period（KMP算法）

Period
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8990 Accepted Submission(s): 4303

Problem 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 AK , 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

//一开始没看懂题意 后来有读懂了 题意就是从str里面寻找前缀zaistr循环出现并且打印其长度
//后来看了别人的博客
说这是对next数组的进一步升华

next  aabaa举例     -1 0 1 0 1      所以i-next[i] 则是回到i前面最近的与i相同的位置

i-next[i] 就是循环体的长度
if(i%(i-next[i])==0) 则说明这个是循环出现的 .

#include<stdio.h>#include<string>#include<string.h>#include<iostream>using namespace std;string str;int nxt[1000007];int len;void Make_Next(){    int i =0,j =-1;    memset(nxt,0,sizeof(nxt));    nxt[0] =-1;    while(str[i])    {        if(j==-1||str[i]==str[j])            nxt[++i] = ++j;        else            j = nxt[j];    }}void kmp(){    for(int i =2;str[i-1];i++)    {        int t = i - nxt[i];        if(i%t==0&&i/t>1)        {            printf("%d %d\n",i,i/t);        }    }}int main(){    int len;    int cnt =1;    while(~scanf("%d",&len),len)    {        cin>>str;        Make_Next();        printf("Test case #%d\n",cnt++);        kmp();        printf("\n");    }    return 0;}