链接：http://poj.org/problem?id=2406
Power Strings
| Time Limit: 3000MS |
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Memory Limit: 65536K |
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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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int M = 1e6 + 5;
char c[M];
int len, fail[M];
void init(){
fail[0] = -1;
int i = 0, j = -1;
while(i < len){
if(j == -1 || c[i] == c[j]) i++, j++, fail[i] = j;
else j = fail[j];
}
}
int main(){
while(scanf("%s", c)){
if(c[0] == ‘.‘)break;
len = strlen(c);
init();
if(len % (len - fail[len]))puts("1");
else printf("%d\n", len / ( len - fail[len]));
}
}
View Code
poj2406Power Strings
原文：https://www.cnblogs.com/EdSheeran/p/9396148.html
