HDU2643(SummerTrainingDay05-P 第二类斯特林数)

Rank

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 433    Accepted Submission(s): 207

Problem Description

Recently in Teddy's hometown there is a competition named "Cow Year Blow Cow".N competitors had took part in this competition.The competition was so intense that the rank was changing and changing.
Now the question is:
How many different ways that n competitors can rank in a competition, allowing for the possibility of ties. 
as the answer will be very large,you can just output the answer MOD 20090126.
Here are the ways when N = 2:
P1 < P2
P2 < P1
P1 = P2 

 

Input

The first line will contain a T,then T cases followed.
each case only contain one integer N (N <= 100),indicating the number of people.

 

Output

One integer pey line represent the answer MOD 20090126.

 

Sample Input

2 2 3

 

Sample Output

3 13

 

Author

teddy

 

Source

百万秦关终属楚

第二类Stirling数 S(p,k)

　　 

S(p,k)的一个组合学解释是：将p个物体划分成k个非空的不可辨别的（可以理解为盒子没有编号）集合的方法数。

k!S(p,k)是把p个人分进k间有差别(如：被标有房号）的房间(无空房）的方法数。

　　 

S(p,k)的递推公式是：S(p,k)=k*S(p-1,k)+S(p-1,k-1) ,1<= k<=p-1

边界条件：S(p,p)=1 ,p>=0    S(p,0)=0 ,p>=1

　　

递推关系的说明：

考虑第p个物品，p可以单独构成一个非空集合，此时前p-1个物品构成k-1个非空的不可辨别的集合，方法数为S(p-1,k-1)；

也可以前p-1种物品构成k个非空的不可辨别的集合，第p个物品放入任意一个中，这样有k*S(p-1,k)种方法。

为n个队员分配k个名次，k的取值为1到n。可将问题转化为将n个球正好放入k的不同的盒子，即为第二类斯特林数，因为k个盒子不同，存在k！个排列，所以为k！× S2[n][k]，答案为sum(i!*S2[n][i]) | 1<=i<=k

//2017-08-05
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long

using namespace std;

const int N = 110;
const int MOD = 20090126;
ll stir2[N][N], ans[N], factorial[N];

void init(){
    factorial[1] = 1;
    for(int i = 2; i < N; i++)
          factorial[i] = (factorial[i-1]*i)%MOD;
    memset(stir2, 0, sizeof(stir2));
    for(int n = 1; n < N; n++){
        stir2[n][1] = 1;
        stir2[n][n] = 1;
        for(int k = 2; k < n; k++){
            stir2[n][k] = stir2[n-1][k-1]+k*stir2[n-1][k];
            stir2[n][k] %= MOD;
        }
    }
}

int main()
{
    int T, n;
    init();
    cin>>T;
    while(T--){
        cin>>n;
        ll ans = 0;
        for(int i = 1; i <= n; i++)
              ans = (ans + factorial[i]*stir2[n][i]) % MOD;
        cout<<ans<<endl;
    }

    return 0;
}