hdu1358 Period

地址：http://acm.hdu.edu.cn/showproblem.php?pid=1358

题目：

Period

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7830    Accepted Submission(s): 3758

Problem Description

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (if there is one) such that the prefix of S with length i can be written as A^K , that is A concatenated K times, for some string A. Of course, we also want to know the period K.

 

 

Input

The input file consists of several test cases. Each test case consists of two lines. The first one contains N (2 <= N <= 1 000 000) – the size of the string S. The second line contains the string S. The input file ends with a line, having the number zero on it.

 

 

Output

For each test case, output “Test case #” and the consecutive test case number on a single line; then, for each prefix with length i that has a period K > 1, output the prefix size i and the period K separated by a single space; the prefix sizes must be in increasing order. Print a blank line after each test case.

 

 

Sample Input

3 aaa 12 aabaabaabaab 0

 

 

Sample Output

Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4

 

 

Recommend

JGShining

 思路：kmp+最小循环节

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int nt[K];
char sa[K],sb[K];

void kmp_next(char *T,int *nt)
{
    nt[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[j]!=T[i])j=nt[j-1];
        if(T[j]==T[i])j++;
        nt[i]=j;
    }
}
int kmp(char *S,char *T,int *nt)
{
    int ans=0;
    kmp_next(T,nt);
    int ls=strlen(S),lt=strlen(T);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j])j=nt[j-1];
        if(S[i]==T[j])j++;
        if(j==lt)
             ans++,j=0;
    }
    return ans;
}
int main(void)
{
    int t,cnt=1;
    while(scanf("%d",&t)&&t)
    {
        scanf("%s",sa);
        kmp_next(sa,nt);
        printf("Test case #%d\n",cnt++);
        for(int i=1;i<t;i++)
        {
            int k=i+1-nt[i];
            if(k==i+1||(i+1)%k!=0)
                continue;
            printf("%d %d\n",i+1,(i+1)/k);
        }
        printf("\n");
    }
    return 0;
}