poj-2406-Power Strings(KMP)
2015-09-20 11:22
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Power Strings
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
Sample Output
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
Time Limit: 3000MS | Memory Limit: 65536K | |
Total Submissions: 38544 | Accepted: 16001 |
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.
/*一个裸的KMP,最后判断一下是否循环*/ #include<stdio.h> #include<string.h> int p[1000010]; char str[1000010]; int len; void getp() { int i=0,j=-1; p[i]=-1; 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)!=EOF) { memset(p,0,sizeof(p)); len=strlen(str); getp(); if(strcmp(str,".")==0) break; if(len%(len-p[len])==0) printf("%d\n",len/(len-p[len])); else printf("1\n"); } return 0; }
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