fzu1901Period II

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

 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <set>
 #include <string>
 using namespace std;

 #define MP make_pair
 #define PB push_back
 typedef long long LL;
 const double eps=1e-;
 const int K=1e6+;
 const int mod=1e9+;

 int nt[K],ans[K];
 char sa[K];
 void kmp_next(char *T,int *next)
 {
     next[]=;
     for(int i=,j=,len=strlen(T);i<len;i++)
     {
         while(j&&T[i]!=T[j]) j=next[j-];
         if(T[i]==T[j])  j++;
         next[i]=j;
     }
 }
 int kmp(char *S,char *T,int *next)
 {
     int ans=;
     int ls=strlen(S),lt=strlen(T);
     for(int i=,j=;i<ls;i++)
     {
         while(j&&S[i]!=T[j]) j=next[j-];
         if(S[i]==T[j])  j++;
         if(j==lt)   ans++;
     }
     return ans;
 }

 int main(void)
 {
     int t,cnt=;cin>>t;
     while(t--)
     {
         scanf("%s",sa);
         kmp_next(sa,nt);
         int len=strlen(sa),k=;
         int tk=len;
         while(nt[tk-]>)
             ans[k++]=len-nt[tk-],tk=nt[tk-];
         ans[k++]=len;
         printf("Case #%d: %d\n",cnt++,k);
         for(int i=;i<k;i++)
             printf("%d%c",ans[i],i!=k-?' ':'\n');
     }
     return ;
 }