首页 > ACM题库 > HDU-杭电 > HDU 3664-Permutation Counting-递推-[解题报告]HOJ
2014
11-30

HDU 3664-Permutation Counting-递推-[解题报告]HOJ

Permutation Counting

问题描述 :

Given a permutation a1, a2, … aN of {1, 2, …, N}, we define its E-value as the amount of elements where ai > i. For example, the E-value of permutation {1, 3, 2, 4} is 1, while the E-value of {4, 3, 2, 1} is 2. You are requested to find how many permutations of {1, 2, …, N} whose E-value is exactly k.

输入:

There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).

输出:

There are several test cases, and one line for each case, which contains two integers, N and k. (1 <= N <= 1000, 0 <= k <= N).

样例输入:

3 0
3 1

样例输出:

1
4

Hint
There is only one permutation with E-value 0: {1,2,3}, and there are four permutations with E-value 1: {1,3,2}, {2,1,3}, {3,1,2}, {3,2,1}

/*
分析:
    递推。
f[i][j]=①(f[i-1][j])+②(j*f[i-1][j])+③(f[i-1][j-1]*(i-j)),
特别的:f[i][0]=1,f[i][i]=0。

                                                     2012-05-10
*/

#include"stdio.h"
__int64 f[1011][1011];
int main()
{
	int i,l;
	int n,k;
	f[1][0]=1;
	f[1][1]=0;
	for(i=2;i<=1000;i++)
	{
		f[i][0]=1;
		f[i][i]=0;
		for(l=1;l<i;l++)
		{
			f[i][l]=f[i-1][l]+l*f[i-1][l]+f[i-1][l-1]*(i-l);
			f[i][l]%=1000000007;
		}
	}


	while(scanf("%d%d",&n,&k)!=-1)	printf("%d\n",f[n][k]);
	
	return 0;
}

参考:http://blog.csdn.net/ice_crazy/article/details/7554933


  1. a是根先忽略掉,递归子树。剩下前缀bejkcfghid和后缀jkebfghicd,分拆的原则的是每个子树前缀和后缀的节点个数是一样的,根节点出现在前缀的第一个,后缀的最后一个。根节点b出现后缀的第四个位置,则第一部分为四个节点,前缀bejk,后缀jkeb,剩下的c出现在后缀的倒数第2个,就划分为cfghi和 fghic,第3部分就为c、c

  2. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮