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

#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;
}

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