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-8;
const int K=1e6+7;
const int mod=1e9+7;

int nt[K],ans[K];
char sa[K];
void kmp_next(char *T,int *next)
{
    next[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[i]!=T[j]) j=next[j-1];
        if(T[i]==T[j])  j++;
        next[i]=j;
    }
}
int kmp(char *S,char *T,int *next)
{
    int ans=0;
    int ls=strlen(S),lt=strlen(T);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j]) j=next[j-1];
        if(S[i]==T[j])  j++;
        if(j==lt)   ans++;
    }
    return ans;
}


int main(void)
{
    int t,cnt=1;cin>>t;
    while(t--)
    {
        scanf("%s",sa);
        kmp_next(sa,nt);
        int len=strlen(sa),k=0;
        int tk=len;
        while(nt[tk-1]>0)
            ans[k++]=len-nt[tk-1],tk=nt[tk-1];
        ans[k++]=len;
        printf("Case #%d: %d\n",cnt++,k);
        for(int i=0;i<k;i++)
            printf("%d%c",ans[i],i!=k-1?' ':'\n');
    }
    return 0;
}