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.
InputInput end with EOF.
Input will be four integers n,r,k,m.We assume that they are all between 1 and 1000.
OutputOutput 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.
题意，现在有1到n，n个数，从中选出r个数，要求选出的数的差至少为k，最后把选出的数放入不大于m个没有差异的盒子里面，问有多少种方案？取模1E9+7
分析：
需要进行两步操作，第一步是要选出r个数，求出共有多少种方案
第二步是在已经有了r个数，那么这r个数的具体值就不重要了，只需求把这r个数放入盒子里的方案数。
最后将两步的结果相乘
对于第一步操作：可以用dp[i][j]表示 数i是第j个数的方案数量
dp[i][j]=(dp[i][j]+dp[z][j-1])%MOD (z表示从1到i-k的数)
但是这样需要三重循环，所需时间超过数据范围
而这里实际上只与前(i-k)个的和有关，所以要用滚动数组的方式进行预处理
第二步：如果是刚好放入m个盒子，就是第二类斯特林数。但这里是放入不大于m个盒子，所以看做刚好放入(1到m)个盒子的第二类斯特林数求和
代码如下：
#include <cstdio> #include <algorithm> #include <iostream> #include <vector> #include <cstring> #include <cmath> using namespace std; typedef long long LL; LL stl[1100][1100]; LL sumstl[1100][1100]; LL dp[1100][1100]; LL d[1100]; LL d2[1100]; const int MOD=1e9+7; void init() { for(int i=1;i<=1000;i++) stl[i][i]=1; for(int i=1;i<=1000;i++) for(int j=1;j<i;j++) stl[i][j]=(stl[i-1][j-1]+j*stl[i-1][j]%MOD)%MOD; for(int i=1;i<=1000;i++) for(int j=1;j<=1000;j++) sumstl[i][j]=(sumstl[i][j-1]+stl[i][j])%MOD; } int main() { LL n,r,k,m,sum,ans; init(); while(scanf("%lld%lld%lld%lld",&n,&r,&k,&m)!=EOF) { sum=0; memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { dp[i][1]=1; d[i]=(d[i-1]+dp[i][1])%MOD; } for(int j=2;j<=r;j++) { for(int i=1;i<=n;i++) { if(i-k>=0) dp[i][j]=(dp[i][j]+d[i-k])%MOD; d2[i]=(d2[i-1]+dp[i][j])%MOD; } for(int z=1;z<=n;z++) d[z]=d2[z]; } for(int i=1;i<=n;i++) sum=(sum+dp[i][r])%MOD; // cout<<sum<<endl; ans=(sum*sumstl[r][m])%MOD; cout<<ans<<endl; } return 0; }