poj 2406 Power Strings【KMP】
来源：程序员人生 发布时间：2014-12-20 09:11:27 阅读次数：2785次
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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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