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
1 #include <cstdio>
2 #include <cstring>
3 const int maxn = 1000010;
4 int p[maxn],ans[maxn];
5 char str[maxn];
6
7 void get_p(int len){
8 p[1] = 0;
9 int j = 0;
10 for(int i = 2;i <= len;i++){
11 while(j > 0 && str[j+1] != str[i])
12 j = p[j];
13 if(str[j+1] == str[i])
14 j++;
15 p[i] = j;
16 }
17 }
18
19 int main(){
20 int nkase;
21 scanf("%d",&nkase);
22 for(int kase = 1;kase <= nkase;kase++){
23 scanf("%s",str+1);
24 int len = strlen(str+1);
25 get_p(len);
26 int t = p[len],cnt = 0;
27 while(t){
28 ans[cnt++] = len-t;
29 t = p[t];
30 }
31 ans[cnt++] = len;
32 printf("Case #%d: %d\n",kase,cnt);
33 for(int i = 0;i < cnt-1;i++)
34 printf("%d ",ans[i]);
35 printf("%d\n",ans[cnt-1]);
36 }
37 return 0;
38 }
FZU 1901 Period II