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poj 2406 Power Strings【KMP】

2024-08-05 06:34:46   219人阅读

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Power Strings
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 33548 Accepted: 13935

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

Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题目翻译意：给一个字符串S长度不超过10^6，求最大的n使得S由n个相同的字符串a连接而成，如："ababab"则由n=3个"ab"连接而成，"aaaa"由n=4个"a"连接而成，"abcd"则由n=1个"abcd"连接而成。

解题思路：假设S的长度为len，则S存在循环子串，当且仅当，len可以被len - next[len]整除，最短循环子串为S[len - next[len]]
利用KMP算法，求字符串的特征向量next，若len可以被len - next[len]整除，则最大循环次数n为len/(len - next[len])，否则为1
#include<cstdio>
#define maxn 1000002
char str[maxn];
int next[maxn],len,s;
void GetNext(){
    int i=0,j=-1;
    next[0]=-1;
    while(str[i]){
        if(j==-1||str[i]==str[j]){
            ++i;++j;next[i]=j;
        }else j=next[j];
    }
    len=i;
}
int main(){
    while(scanf("%s",str)==1){
        if(str[0]=='.') break;
        GetNext();
        s=len-next[len];
        if(len%s==0) printf("%d\n",len/s);
        else printf("1\n");
    }return 0;
}


poj 2406 Power Strings【KMP】


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