POJ-2406 Power Strings (KMP)
2018-03-18 14:52
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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.Ouput
For each s you should print the largest n such that s = a^n for some string a.Sample Input
abcdaaaa
ababab
.
Sample Output
14
3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.思路
这几天看KMP看的云里雾里……好像有点理解了它的思想了。关于这道题目,首先根据KMP的思想求出给定字符串对应的next数组,可知len-next[len]为该字符串最短循环子串的长度,若len%(len-next[len]) == 0则存在n符合题意,否则不存在。关于这个思想的理解我是参考了博客POJ 2406 Power Strings (KMP) 总觉得还要再好好理解理解KMP啊(雾代码
#include<iostream> #include<cstdlib> #include<stdio.h> #include<string.h> using namespace std; char str[1000010]; int len, ans, next[1000010]; void getNext() { int i = 0, j = -1; next[0] = -1; while(i < len) { if (j == -1 || str[i] == str[j]) { i++; j++; next[i] = str[i] != str[j] ? j : next[j]; } else j = next[j]; } } int main() { while(scanf("%s", str)) { if(str[0] == '.') break; len = strlen(str); getNext(); if(len % (len - next[len]) == 0) ans = len / (len - next[len]); else ans = 1; cout << ans << endl; } }
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