Problem 1901 Period II【KMP--next数组】

题目：

For each prefix with length P of a given string S,if

S[i]=S[i+P] for i in [0..SIZE(S)-p-1],

then the prefix is a “period” of S. We want to all the periodic prefixs.

Input

Input contains multiple cases.

The first line contains an integer T representing the number of cases. Then following T cases.

Each test case contains a string S (1 <= SIZE(S) <= 1000000),represents the title.S consists of lowercase ,uppercase letter.

Output

For each test case, first output one line containing "Case #x: y", where x is the case number (starting from 1) and y is the number of periodic prefixs.Then output the lengths of the periodic prefixs in ascending order.

Sample Input
4
ooo
acmacmacmacmacma
fzufzufzuf
stostootssto
Sample Output
Case #1: 3
1 2 3
Case #2: 6
3 6 9 12 15 16
Case #3: 4
3 6 9 10
Case #4: 2
9 12

题目大意:

求所有可以由循环得到整个字符串的前缀的长度【循环节的最后允许欠缺】；

即求出所有满足s[i] == s[i+p] ( 0 < i+p < len )的 p ;

解题思路：

如果目标串的当前字符i在匹配到模式串的第j个字符时失配，那么我们可以让i试着去匹配next(j)
对于模式串str，next数组的意义就是：
   如果next(j)=t,那么str[1…t]=str[len-t+1…len]
  我们考虑next(len)，令t=next(len);
next(len)有什么含义？
   str[1…t]=str[len-t+1…len]
那么，长度为len-next(len)的前缀显然是符合题意的。
接下来我们应该去考虑谁？
   t=next( next(len) );
   t=next( next (next(len) ) );
  一直下去直到t=0,每个符合题意的前缀长是len-t

例：

aaabaaa：--> P = 4 <---> next[7] = 3

aaabaaa：--> P = 5 <---> next[3] = 2

aaabaaa：--> P = 6 <---> next[2] = 1

aaabaaa：--> P = 7 <---> next[1] = 0

实现代码：

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MM=1000005;

int ne[MM],a[MM];
char mo[MM];
int lm;
void Get_next(){
    int i=0,j=-1;
    ne[0]=-1;
    while(i<lm){
        while(j!=-1&&mo[i]!=mo[j])
            j=ne[j];
        ne[++i]=++j;
    }
}
int main(){
    int ca,cas=1;
    scanf("%d",&ca);
    while(ca--){
        scanf("%s",mo);
        int k=0;
        lm=strlen(mo);
        a[k++]=lm;
        memset(ne,0,sizeof(ne));
        Get_next();
        int tmp=ne[lm];
        while(tmp){
            a[k++]=lm-tmp;
            tmp=ne[tmp];
        }
        printf("Case #%d: %d\n",cas++,k);
        for(int i=1;i<k;i++)
            printf("%d ",a[i]);
        printf("%d\n",a[0]);
    }
    return 0;
}