【第一类斯特林数】HDU_4372_ Count the Buildings

Count the Buildings

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2051    Accepted Submission(s): 676

Problem Description

There are N buildings standing in a straight line in the City, numbered from 1 to N. The heights of all the buildings are distinct and between 1 and N. You can see F buildings when you standing in front of the first building and looking forward, and B buildings when you are behind the last building and looking backward. A building can be seen if the building is higher than any building between you and it.
Now, given N, F, B, your task is to figure out how many ways all the buildings can be.

 

Input

First line of the input is a single integer T (T<=100000), indicating there are T test cases followed.
Next T lines, each line consists of three integer N, F, B, (0<N, F, B<=2000) described above.

 

Output

For each case, you should output the number of ways mod 1000000007(1e9+7).

 

Sample Input

23 2 23 2 1

 

Sample Output

21

 

Source

2012 Multi-University Training Contest 8

 

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zhuyuanchen520

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn=2005;const int mod=1e9+7;LL dp[maxn][maxn],c[maxn][maxn];void Init(int n,int k){    c[0][0]=1;    for(int i=1;i<=n;i++){        c[i][0]=1;         for(int j=1;j<=i;j++)            c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;    }    for(int i=0;i<=n;i++)        dp[0][i]=0;    dp[1][1]=1;    for(int i=2;i<=n;i++)        for(int j=1;j<=i;j++)            dp[i][j]=(dp[i-1][j-1]%mod+(i-1)%mod*dp[i-1][j]%mod)%mod;}int main(){    int t,N,F,B;    Init(2000,2000);    scanf("%d",&t);    while(t--){        scanf("%d%d%d",&N,&F,&B);        if(F+B-1>N){            puts("0");            continue;        }        LL ans=dp[N-1][F-1+B-1]*c[F-1+B-1][F-1]%mod;        printf("%I64d\n",ans);    }    return 0;}