poj 2406 Power Strings【KMP】

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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1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:假定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,求字符串的特点向量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=⑴;
next[0]=⑴;
while(str[i]){
if(j==⑴||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
",len/s);
else printf("1
");
}return 0;
}
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波比源码 » poj 2406 Power Strings【KMP】

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