[2406]Power Strings (POJ) KMP



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Language: Default

Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 32710		Accepted: 13635

Description

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
char a[1000010],b[1000010];
int next[1000010];
int n,num=1;
void get_next()//求next数组
{
    int len=strlen(a);
    int i=0,j=-1;
    next[0]=-1;
    while(i<len)
    {
        if(j==-1||a[i]==a[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else
            j=next[j];
    }
}
int main()
{
    while(gets(a))
    {
        if(a[0]=='.')
            break;
        int len=strlen(a);
        get_next();
        int n=len-next[len];//n为循环节的长度
        if(len%n==0)
            printf("%d\n",len/n);
        else
            printf("1\n");
    }
    return 0;
}

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