POJ 2406:Power Strings
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Power Strings
Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 41252 Accepted: 17152
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
abcdaaaaababab.
Sample Output
143
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.
题意:
给出的一个字符串某个字串链接N次得出的,求最大N;
思想:
KMP匹配算法之next[];模式串结尾指针回溯最小移动次数即最小字串长度,N=s.length%(s.length-next[s.length]);
#include <stdio.h>#include <string.h>char s[1000005];int num[1000005];int main(){ while(scanf("%s",s),s[0]!='.') { int j=-1; num[0]=-1; int n=strlen(s); for(int i=0; i<n;) { if(j==-1||s[i]==s[j])num[++i]=++j; else j=num[j]; } if(n%(n-num[n])==0) printf("%d\n",n/(n-num[n])); else printf("1\n"); } return 0;}
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