地址：http://acm.fzu.edu.cn/problem.php?pid=1901
题目：
Problem 1901 Period II
Accept: 442 Submit: 1099
Time Limit: 1000 mSec Memory Limit : 32768 KB
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
Source
FOJ有奖月赛-2010年05月思路：next数组
1 #include <cstdio>
2 #include <cstring>
3 #include <iostream>
4 #include <set>
5 #include <string>
6 using namespace std;
7
8 #define MP make_pair
9 #define PB push_back
10 typedef long long LL;
11 const double eps=1e-8;
12 const int K=1e6+7;
13 const int mod=1e9+7;
14
15 int nt[K],ans[K];
16 char sa[K];
17 void kmp_next(char *T,int *next)
18 {
19 next[0]=0;
20 for(int i=1,j=0,len=strlen(T);i<len;i++)
21 {
22 while(j&&T[i]!=T[j]) j=next[j-1];
23 if(T[i]==T[j]) j++;
24 next[i]=j;
25 }
26 }
27 int kmp(char *S,char *T,int *next)
28 {
29 int ans=0;
30 int ls=strlen(S),lt=strlen(T);
31 for(int i=0,j=0;i<ls;i++)
32 {
33 while(j&&S[i]!=T[j]) j=next[j-1];
34 if(S[i]==T[j]) j++;
35 if(j==lt) ans++;
36 }
37 return ans;
38 }
39
40
41 int main(void)
42 {
43 int t,cnt=1;cin>>t;
44 while(t--)
45 {
46 scanf("%s",sa);
47 kmp_next(sa,nt);
48 int len=strlen(sa),k=0;
49 int tk=len;
50 while(nt[tk-1]>0)
51 ans[k++]=len-nt[tk-1],tk=nt[tk-1];
52 ans[k++]=len;
53 printf("Case #%d: %d
",cnt++,k);
54 for(int i=0;i<k;i++)
55 printf("%d%c",ans[i],i!=k-1?' ':'
');
56 }
57 return 0;
58 }