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POJ-2406-Power Strings

无愠色 2023-02-02 阅读 97



Power Strings

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 96   Accepted Submission(s) : 34


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).


 




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


题目分析:

意思是有多少个子串构成比如说  aaaa = a^4   ababab=(ab)^3    abcd=(abcd)^1    应该明白了吧!也就是这个字符串能被他的最短子串构成   由kmp ,next表的特点知如果满足  len%(len-p[len])==0  就能由子串构成且子串个数就是 len/(len-p[len])




#include<cstdio>
#include<cstring>
const int M = 1000000+10;
char str[M];
int p[M],a[M],len;
void getp()
{
int i=0,j=-1;
p[i]=j;
while(i<len)
{
if(j==-1||str[i]==str[j])
{
i++;j++;
p[i]=j;
}
else j=p[j];
}
}
int main()
{
while(~scanf("%s",str)&&str[0]!='.')
{
len=strlen(str);
getp();
if(len%(len-p[len])==0)
printf("%d\n",len/(len-p[len]));
else
printf("1\n");
}
return 0;
}



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