## HDU/HDOJ 4045 BUPT 216 Machine scheduling 2011ACM北京网络赛 F题

2018年01月20日 ⁄ 综合 ⁄ 共 2301字 ⁄ 字号 评论关闭

Machine scheduling
 Accept:8 Submit:35 Time Limit:2000MS Memory Limit:65536KB
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.

s[i][j]=s[i-1][j]+s[i][j-1]

s[n][m]=m*s[n-1][m]+s[n-1][m-1]

ans=s[n]+s[n]+...+s[n][m]

BUPT上面交的话，记得把__int64改为long long，%I64d改为%lld

```#include<stdio.h>
#include<string.h>
#include<algorithm>

using namespace std;

typedef __int64 ll;

ll sum1;
ll sum2;
ll mod=1000000007;

void init()
{
ll i,j;
for(i=1;i<=1000;i++)
{
sum1[i]=1;
sum1[i]=i;
}
for(i=2;i<=1000;i++)
for(j=2;j<=1000;j++)
sum1[i][j]=(sum1[i-1][j]+sum1[i][j-1])%mod;
for(i=1;i<=1000;i++)
{
sum2[i]=1;
sum2[i][i]=1;
}
for(i=2;i<=1000;i++)
for(j=2;j<=i;j++)
sum2[i][j]=(j*sum2[i-1][j]+sum2[i-1][j-1])%mod;
}

ll cal(ll n,ll m)
{
ll i,sum=0;
for(i=1;i<=m;i++)
sum=(sum+sum2[n][i])%mod;
return sum;
}

int main()
{
ll n,r,k,m,tmp,ans;
init();
while(scanf("%I64d%I64d%I64d%I64d",&n,&r,&k,&m)!=EOF)
{
tmp=k*(r-1)+1;
if(tmp>n)
{
printf("0\n");
continue;
}
tmp=n-k*(r-1);
ans=cal(r,m)*sum1[r][tmp]%mod;
printf("%I64d\n",ans);
}
return 0;
}```