POJ 题目2406 Power Strings（KMP）

最新推荐文章于 2022-01-29 11:49:34 发布

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最新推荐文章于 2022-01-29 11:49:34 发布

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本文深入探讨了一种在面对大量输入时，高效处理字符串操作的方法。通过利用预处理技巧和动态规划思想，该算法能够在有限时间内解决字符串幂次方问题，特别针对输入长度可达百万级的挑战，提出了一套有效的解决方案。

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Power Strings

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 33655		Accepted: 13983

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

ac代码

#include<stdio.h>
#include<string.h>
__int64 a[1010][110],b[1010][110],inf=1;
int main()
{
	int n,k,i;
	for(i=0;i<18;i++)
		inf*=10;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		int i,j;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(i=0;i<=k;i++)
			a[0][i]=1;
		for(j=1;j<=k;j++)
		{
			for(i=1;i<=n;i++)
			{
				if(i<j)
				{
					a[i][j]=a[i][j-1];
					b[i][j]=b[i][j-1];
				}
				else
				{
					a[i][j]=(a[i-j][j]+a[i][j-1])%inf;
					b[i][j]=b[i-j][j]+b[i][j-1]+(a[i-j][j]+a[i][j-1])/inf;
				}
			}
		}
		if(b[n][k])
			printf("%I64d",b[n][k]);
		printf("%I64d\n",a[n][k]);
	}
}