FZU 1011 Power Strings(KMP匹配演算法)
Problem 1011 Power Strings
Accept: 914 Submit: 2751
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem 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).
Each test case is a line of input representing s, a string of printable characters. For each s you should print the largest n such that s = a^n for some string a. 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.
Sample Input
abcd
aaaa
ababab
.
Sample Output
1
4
3
本題為簡單的KMP匹配演算法,,是我在KMP這個演算法中做的第一道題.
下面是我的AC程式碼
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char str[1000005];
int next[1000005];
int kmp()
{
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];
}
return j;
}
int main()
{
while(gets(str)&&str[0]!='.')
{
int len=strlen(str);
int r=len-kmp();
if(len%r==0)
{
printf("%d\n",len/r);
}
else printf("1\n");
}
return 0;
}