poj 2406 Power Strings（KMP入门，next函数理解）

Power Strings

Time Limit: 3000MS Memory Limit: 65536K

Total Submissions: 37685 Accepted: 15590

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
Hint

This problem has huge input, use scanf instead of cin to avoid time limit exceed.

Source

Waterloo local 2002.07.01

基础题目，只要是理解next函数。

 #include<iostream>

 #include<vector>

 #include <cstdio>

 #include <cstring>

 #include <cstdlib>

 #include <math.h>

 #include<algorithm>

 #define ll long long

 #define eps 1e-8

 using namespace std;

 int nexts[];

 char b[];

 void pre_nexts(int m)

 {

     memset(nexts,,sizeof(nexts));

     int j = ,k = -;

     nexts[] = -;

     while(j < m)

     {

         if(k == - || b[j] == b[k])  nexts[++j] = ++k;

         else k = nexts[k];

     }

 }

 int main(void)

 {

     while(scanf("%s",b),b[] != '.')

     {

         int n = (int)strlen(b);

         pre_nexts(n);

         if(n % (n-nexts[n]) ==  && n/(n - nexts[n]) > ) printf("%d\n",n/(n-nexts[n]));

         //根据next函数的最后一个匹配数，算出循环节点～

         else printf("1\n");

     }

     return ;

 }

巴特西

poj 2406 Power Strings（KMP入门，next函数理解）

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Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 37685		Accepted: 15590