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>
#include <algorithm>

using namespace std;

typedef long long int LL;
const long long int MOD=1000000007;

LL s[5050][5050],c[5050][5050];

void init()
{
    for(int i=0;i<5050;i++) c[i][i]=c[i][0]=1LL;
    for(int i=2;i<5050;i++)
    {
        for(int j=1;j<i;j++)
        {
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%MOD;
        }
    }

    s[0][0]=1;
    for(int i=1;i<5050;i++)
    {
        for(int j=1;j<=i;j++)
        {
            s[i][j]=((i-1)*s[i-1][j]%MOD+s[i-1][j-1])%MOD;
        }
    }
}

void solve(int n,int left,int right)
{
    LL ans=0;
    for(int i=left;i<=n-right+1;i++)
    {
        ans=(ans+((c[n-1][i-1]*s[i-1][left-1])%MOD)*s[n-i][right-1]%MOD)%MOD;
    }
    printf("%I64d\n",ans%MOD);
}

int main()
{
    int n,left,right;
    init();
while(scanf("%d%d%d",&n,&left,&right)!=EOF&&n&&left&&right) solve(n,left,right);
    return 0;
}