FZU Problem 1901 Period II

Problem Description

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

题意：给你一个字符串str，对于每个str长度为p的前缀，如果str[i]==str[p+i](p+i<len)，那么我们认为它是一个periodic prefixs.求所有满足题意的前缀的长度p。
知识点：KMP算法、对next数组的理解
KMP算法中next数组的含义是什么？
next数组：失配指针
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

#include <cstdio>
#include <cstring>
const int maxn = 1000010;
int p[maxn],ans[maxn];
char str[maxn];

void get_p(int len){
p[1] = 0;
int j = 0;
for(int i = 2;i <= len;i++){
while(j > 0 && str[j+1] != str[i])  j = p[j];
if(str[j+1] == str[i])  j++;
p[i] = j;
}
}

int main(){
int nkase;
scanf("%d",&nkase);
for(int kase = 1;kase <= nkase;kase++){
scanf("%s",str+1);
int len = strlen(str+1);
get_p(len);
int t = p[len],cnt = 0;
while(t){
ans[cnt++] = len-t;
t = p[t];
}
ans[cnt++] = len;
printf("Case #%d: %d\n",kase,cnt);
for(int i = 0;i < cnt-1;i++)  printf("%d ",ans[i]);
printf("%d\n",ans[cnt-1]);
}
return 0;
}