hdu 4045 Machine scheduling 斯特林数+插板法

最新推荐文章于 2025-06-14 21:42:16 发布

Lionel_D

最新推荐文章于 2025-06-14 21:42:16 发布

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分类专栏：数论数据结构排列组合文章标签： hdu4045 Machine scheduling 组合数学斯特林数

本文链接：https://blog.youkuaiyun.com/lionel_d/article/details/45565285

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本博客探讨了百度工程师在使用机器处理大量数据时的机器调度问题，旨在找到在不重复分配方案的情况下，机器可以连续使用的最大天数。通过输入机器数量、选择机器数量、最小工作间隔和组数，输出连续使用天数模运算结果。

Machine scheduling

Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1069 Accepted Submission(s): 400

Problem Description

A Baidu’s engineer needs to analyze and process large amount of data on machines every day. The machines are labeled from 1 to n. On each day, the engineer chooses r machines to process data. He allocates the r machines to no more than m groups ,and if the difference of 2 machines' labels are less than k,they can not work in the same day. Otherwise the two machines will not work properly. That is to say, the machines labeled with 1 and k+1 can work in the same day while those labeled with 1 and k should not work in the same day. Due to some unknown reasons, the engineer should not choose the allocation scheme the same as that on some previous day. otherwise all the machines need to be initialized again. As you know, the initialization will take a long time and a lot of efforts. Can you tell the engineer the maximum days that he can use these machines continuously without re-initialization.

Input

Input end with EOF.
Input will be four integers n,r,k,m.We assume that they are all between 1 and 1000.

Output

Output the maxmium days modulo 1000000007.

Sample Input

5 2 3 2

Sample Output

6

Hint
Sample input means you can choose 1 and 4,1 and 5,2 and 5 in the same day.
And you can make the machines in the same group or in the different group.
So you got 6 schemes.
1 and 4 in same group,1 and 4 in different groups.
1 and 5 in same group,1 and 5 in different groups.
2 and 5 in same group,2 and 5 in different groups.
We assume 1 in a group and 4 in b group is the same as 1 in b group and 4 in a group.

传送门：http://blog.youkuaiyun.com/acm_cxlove/article/details/7857785

代码：

#include <stdio.h>
#define MOD 1000000007
int c[2010][2001] = {1} , stir[1010][1010] = {1} ;
int min(int a , int b)
{
	return a>b?b:a ;
}
int main()
{
	for(int i = 0 ; i <= 2000 ; ++i)
	{
		c[i][0] = c[i][i] = 1 ;
		for(int j = 1 ; j < i ; ++j)
		{
			c[i][j] = (c[i-1][j-1]+c[i-1][j])%MOD ;
		}
	}
	for(int i = 1 ; i <= 1000 ; ++i)
	{
		stir[i][0] = 0 ;
		stir[i][i] = 1 ;
		for(int j = 1 ; j < i ; ++j)
		{
			stir[i][j] = ((long long)j*stir[i-1][j]+stir[i-1][j-1])%MOD ;
		}
	}
	int n,r,k,m ;
	while(~scanf("%d%d%d%d",&n,&r,&k,&m))
	{
		int rem = n-(r-1)*k-1 ;
		if(rem<0)
		{
			puts("0") ;
		}
		else
		{
			long long sum = 0 ;
			int len = min(r,m) ;
			for(int i = 1 ; i <= len ; ++i)
			{
				sum = (sum+stir[r][i])%MOD ;
			}
			printf("%d\n",c[rem+r][r]*sum%MOD) ;
		}
	}
	return 0 ;
}

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