C. White, Black and White Again
time limit per test
memory limit per test
input
output
Polycarpus is sure that his life fits the description: "first there is a white stripe, then a black one, then a white one again". So, Polycarpus is sure that this rule is going to fulfill during the next n days. Polycarpus knows that he is in for w good events and b
What is the number of distinct ways this scenario can develop over the next n days if Polycarpus is in for a white stripe (a stripe that has good events only, the stripe's length is at least 1 day), the a black stripe (a stripe that has not-so-good events only, the stripe's length is at least 1 day) and a white stripe again (a stripe that has good events only, the stripe's length is at least 1 day). Each of n
Note that even the events of the same type are distinct from each other. Even if some events occur on the same day, they go in some order (there are no simultaneous events).
Write a code that prints the number of possible configurations to sort the events into days. See the samples for clarifications on which scenarios should be considered distinct. Print the answer modulo 1000000009 (109 + 9).
Input
The single line of the input contains integers n, w and b (3 ≤ n ≤ 4000, 2 ≤ w ≤ 4000, 1 ≤ b ≤ 4000) — the number of days, the number of good events and the number of not-so-good events. It is guaranteed that w + b ≥ n.
Output
Print the required number of ways modulo 1000000009 (109 + 9).
Examples
Input
3 2 1
Output
2
Input
4 2 2
Output
4
Input
3 2 2
Output
4
Note
We'll represent the good events by numbers starting from 1 and the not-so-good events — by letters starting from 'a'. Vertical lines separate days.
In the first sample the possible ways are: "1|a|2" and "2|a|1". In the second sample the possible ways are: "1|a|b|2", "2|a|b|1", "1|b|a|2" and "2|b|a|1". In the third sample the possible ways are: "1|ab|2", "2|ab|1", "1|ba|2" and "2|ba|1".
【分析】
切水题系列...
枚举坏事发生了多少天，根据隔板法方案数，O(1)计算即可。
注意要乘阶乘，因为一些事在同一天发生的顺序要考虑到。
【代码】
//codeforces 306C White, Black and White Again
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define fo(i,j,k) for(int i=j;i<=k;i++)
using namespace std;
const int mod=1e9+9;
const int mxn=4005;
int n,w,b;
ll ans,res;
ll c[mxn][mxn],fac[mxn];
inline void init()
{
int tmp=max(n,max(w,b));
fo(i,0,tmp) c[i][0]=1;
fo(i,1,tmp)
fo(j,1,i)
c[i][j]=(c[i-1][j]+c[i-1][j-1])%mod;
fac[0]=1;
fo(i,1,max(w,b)) fac[i]=(i*fac[i-1])%mod;
}
int main()
{
scanf("%d%d%d",&n,&w,&b);
init();
fo(i,1,b)
{
ll res=n-i-1;
if(!res) break;
res=res*c[b-1][i-1]%mod;
res=res*fac[b]%mod;
res=res*c[w-1][n-i-1]%mod;
res=res*fac[w]%mod;
ans=(ans+res)%mod;
}
printf("%I64d\n",ans);
return 0;
}
//100 200 300
