SERC 2013 E.Skyscrapers

http://acm.hunnu.edu.cn/online/?action=problem&type=show&id=11445

E:   Skyscrapers

Skyscrapers  is a pencil puzzle. It’s played on a square  nxn  grid. Each cell of the grid has a 
building.  Each  row,  and  each  column,  of  the  grid  must  have  exactly  one  building  of 
height  1,  height  2, height 3,  and  so  on,  up to height  n. There  may be  numbers  at  the 
beginning and end of each row, and each column. They indicate how many buildings can 
be seen from that vantage point, where taller buildings obscure shorter buildings. In the 
game, you are given the numbers along the outside of the grid, and you must determine 
the heights of the buildings in each cell of the grid.
Consider  a  single  row  of  a  puzzle  of  size  nxn.  If  we  know  how  many  buildings  can  be 
seen from the left, and from the right, of the row, how many different ways are there of 
populating that row with buildings of heights 1..n? 
Input
There will be several  test cases in the input. Each test case consists of  three integers n a 
single line:  n  (1≤n≤5,000),  left  (1≤left≤n), and  right  (1≤right≤n), where  n  is the size of 
the row, and  left  and  right  are the number of buildings that can be seen from the left 
and right, respectively. The Input will end with a line with three 0s.
Output
For each  test case, print a single  integer indicating the number of permutations which 
satisfy  the  constraints,  modulo  1,000,000,007  (that’s  not  a  misprint,  the  last  digit  is  a 
seven). Output no extra spaces, and do not separate answers with blank lines.
Sample Input
3 2 2
4 1 2
0 0 0
Sample Output
2

2

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define ll long long
const ll mod=1e9+7;
#define N 5005
ll c[5055][5055],dp[5055][5055];
int n,l,r;
void init()
{
    for(int i=0;i<N;i++)c[i][0]=1,c[i][i]=1;
    for(int i=2;i<N;i++)
        for(int j=1;j<i;j++)
        c[i][j]=c[i-1][j-1]+c[i-1][j],c[i][j]%=mod;
    dp[0][0]=1;
    for(int i=1;i<N;i++)
        for(int j=1;j<=i;j++)
        dp[i][j]=(i-1)*dp[i-1][j]%mod+dp[i-1][j-1],dp[i][j]%=mod;
}

int main()
{
    init();
    while(cin>>n>>l>>r&&n)
    {
        ll ans=0;
        for(int i=l;i<=n-r+1;i++)
        {
            ans+=c[n-1][i-1]*dp[i-1][l-1]%mod*dp[n-i][r-1]%mod,ans%=mod;
        }
        cout<<ans<<endl;
    }
}