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
题意:
找出字符串中某一段子串的循环次数。。。。
ac代码:
#include<stdio.h>
#include<cstring>
const int MAXN=1e6+2;
int next[MAXN];
char s[MAXN];
int m;
void get_next()
{
int i=1;
int j=0;
next[0]=0;
while(i<m){
if(s[i]==s[j])
{
next[i]=++j;
i++;
}
else if(!j){
i++;
}
else{
j=next[j-1];
}
}
i=m-j;
if(m%i==0)
printf("%d\n",m/i);
else
printf("1\n");
}
int main()
{
while(scanf("%s",s)&&s[0]!='.')
{
m=strlen(s);
get_next();
}
return 0;
}
报道。。。打卡