http://poj.org/problem?id=2406
Power Strings
| Time Limit: 3000MS |
|
Memory Limit: 65536K |
| Total Submissions:66981 |
|
Accepted: 27644 |
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.
#include<cstdio>
#include<cstring>
const int maxn=1e7+3;
int len;
int next[maxn];
char str[maxn];
void getnext(){
int k=-1;
int j=0;
next[0]=-1;
while(j<len){
if(k==-1||str[j]==str[k]){
k++;
j++;
next[j]=k;
}else{
k=next[k];
}
}
return;
}
int main(){
while(scanf("%s",str)&&str[0]!=‘.‘){
len=strlen(str);
getnext();
int tmp=len-next[len];
if(len%tmp!=0){
printf("1\n");
}else{
printf("%d\n",len/tmp);
}
}
return 0;
}
poj 2406
原文：https://www.cnblogs.com/qqshiacm/p/11443798.html
