Machine scheduling
Time Limit: 5000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1000 Accepted Submission(s): 363
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.
Source
The 36th ACM/ICPC
Asia Regional Beijing Site —— Online Contest
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解题思路：
这题考的是排列组合
（1）首先，要从n个元素编号1~n中选出r个元素来，使得任意两个元素编号相差>=k
解法：先按照 1 k+1 2*k+1 .... (r-1)*k+1 也就是刚好 间隔 k个排列下来
那么 总共n个元素，剩余 n-(r-1)*k-1个元素可以看成空格，极为space,将它们插入这r个元素的r+1个空档中，
这样就能保证枚举了素有的方法数，切满足任意两个元素编号相差>=k的要求
所以这个问题简化为：将space个元素分配到r+1个区域中，可以为空
答案为：stir2[space][r+1],第二类斯特林数
（2）然后，再将选取的r个元素分为不超过g组，枚举想要的组数即可
只需要枚举对应的组数， 1~min(r,g)
转化问题：将n个元素最多分为m个集合，不为空的方案法,插板法，答案为：c[n+m-1][m-1]
解题代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
const int maxn=1010;
int n,r,m,g;
ll c[2*maxn][2*maxn],stir2[maxn][maxn];
void ini(){
for(int i=1;i<2*maxn;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<maxn;i++){
stir2[i][0]=0;
stir2[i][i]=1;
for(int j=1;j<i;j++)
stir2[i][j]=((ll)j*stir2[i-1][j]+stir2[i-1][j-1])%mod;
}
}
ll get1(ll n,ll m){//将n个元素最多分为m个集合，不为空的方案法。
return c[n+m-1][m-1];
}
ll get2(ll n,ll m){//将n个元素分配到m个区域中，可以为空。
return stir2[n][m];
}
void solve(){
int space=n-(r-1)*m-1;
if(space<0){
printf("0\n");
return ;
}
ll ans=0;
//将r个元素分为i组
for(int i=1;i<=min(r,g);i++){
ans=(ans+get2(r,i))%mod;
}
ans=( ans*get1(space,r+1) )%mod;
cout<<ans<<endl;
}
int main(){
ini();
while(scanf("%d%d%d%d",&n,&r,&m,&g)!=EOF){
solve();
}
return 0;
}