Power Strings
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 50546 | Accepted: 21082 |
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
题解:假设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-1]整除,最短循环子串为S[len - next[len-1]]
根据算法的不同,这里的next[len-1]代表整个字符串(即算上最后一位字符)的最大匹配值。
若不能整除,输出1;
代码如下:
#include<cstdio>
#include<cstring>
char P[1000005];
int Next[1000005];
int main()
{
while(~scanf("%s",P)){
if(P[0]=='.') break;
int i,k=0;
memset(Next,0,sizeof(Next));
int n=strlen(P);
Next[0]=0;
for(i=1;i<n;i++){
while(k>0&&P[i]!=P[k])
k=Next[k-1];
if(P[i]==P[k])
k++;
Next[i]=k;
}
if(n%(n-Next[n-1])==0)
printf("%d\n",n/(n-Next[n-1]));
else
printf("1\n");
}
return 0;
}