poj——2406Power Strings（kmp专练）

最新推荐文章于 2022-01-29 11:49:34 发布

最新推荐文章于 2022-01-29 11:49:34 发布 · 286 阅读

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#c++ #poj #string #kmp

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本文介绍了一种通过计算字符串的循环节来确定其最大重复次数的方法。利用KMP算法的next数组，找到输入字符串的循环节长度，进而计算出最大重复次数。

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Power Strings
Time Limit: 3000MS Memory Limit: 65536K
Total Submissions: 41658 Accepted: 17323
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

解题思路：找出循环节用总长度除一下得出循环次数如果除不尽说明循环一次

#include<iostream>
#include<string>
#include<cstring>
using namespace std;
int nexta[4000100];
void getnext(string n)
{
	int i=0,j=-1;
	nexta[0]=-1;
	while(i<n.size())
	{
		if(j==-1||n[i]==n[j])
		{
			++i;
			++j;
			nexta[i]=j;
		}
		else j=nexta[j];
	}
}
int main()
{
	string n;
	long long a,k,i;
	ios::sync_with_stdio(false);
	while(cin>>n&&n!=".")
	{
		getnext(n);
		i=n.size();
		k=i-nexta[i];//循环节 
		if(i%k==0)
		{
			cout<<i/k<<endl;
		}
		else cout<<"1"<<endl;
	}
	return 0;
}