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HDU 4372 Count the Buildings

2018-06-17 21:15:08来源：未知阅读 ()

Count the Buildings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2521 Accepted Submission(s): 817

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

2 3 2 2 3 2 1

Sample Output

2 1

Source

2012 Multi-University Training Contest 8

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zhuyuanchen520 | We have carefully selected several similar problems for you: 4059 2819 1730 2176 1850

实在搞不懂HDU的脾气啊。

为什么交G++会RE？？？？

这道题的话。

不管是从左边看还是从右边看，视线总是会被中间最高的给挡住

所以我们把左边和右边分组来看。

对于某一边，我们确定出能够看见的楼房，那么不能够看见的楼房就可以任意排列

我们把能看见的楼房，与下一个能看到的楼房（不包括下一个楼房）之间的楼看为一组

那么组内的元素就可以任意排，假设一组有$n$个楼房，那么它们自由排列的方案数就是$(n-1)!$

这会让你联想的到什么？

哈哈没错，第一类斯特灵数（神TM能想到）

这样我们就可以首先对所有的楼房进行分组，然后再让他们自由排列

一共有$F-1+B-1$组

#include<iostream>
#include<cstdio>
#define LL long long 
using namespace std;
const LL MAXN=2*1e6+10;
LL T,N,F,B;
LL mod=1000000007;
LL c[2011][2011],s[2011][2011];
int main()
{
    for(LL i=0;i<=2010;i++)
    {
        c[i][0]=1;
        c[i][i]=1;
        s[i][0]=0;//无法频出环 
        s[i][i]=1;//只能拼出一个环 
        for(LL j=1;j<i;j++)
            c[i][j]=(c[i-1][j-1]%mod+c[i-1][j]%mod)%mod,
            s[i][j]=(s[i-1][j-1]%mod+(i-1)%mod*s[i-1][j]%mod)%mod;
    }
    scanf("%I64d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d",&N,&F,&B);
        printf("%I64d\n",(c[F-1+B-1][F-1]%mod*s[N-1][F-1+B-1]%mod)%mod);
    }
    return 0;
}