bzoj 3622: 已经没有什么好害怕的了 （容斥原理+DP）

题目描述

传送门

题解

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#define N 2003
#define p 1000000009
#define LL long long
using namespace std;
int n,k;
LL f

,jc
,inv
;
int a
,b
,pos
;
LL calc(int n,int m)
{
return jc
*inv[m]%p*inv[n-m]%p;
}
LL quickpow(LL num,int x)
{
LL ans=1; LL base=num%p;
while (x) {
if (x&1) ans=ans*base%p;
x>>=1;
base=base*base%p;
}
return ans;
}
int main()
{
freopen("a.in","r",stdin);
freopen("my.out","w",stdout);
scanf("%d%d",&n,&k);
for (int i=1;i<=n;i++) scanf("%d",&a[i]);
for (int i=1;i<=n;i++) scanf("%d",&b[i]);
jc[0]=1;
for (int i=1;i<=n;i++) jc[i]=jc[i-1]*i%p;
for (int i=0;i<=n;i++) inv[i]=quickpow(jc[i],p-2)%p;
sort(a+1,a+n+1);
sort(b+1,b+n+1);
int l=1;
for (int i=1;i<=n;i++) {
while (b[l]<a[i]&&l<=n) l++;
pos[i]=l-1;
}
f[0][0]=1;
for (int i=1;i<=n;i++)
for (int j=0;j<=n;j++) {
f[i][j]=f[i-1][j];
if (j-1>=0)
f[i][j]=(f[i][j]+(LL)f[i-1][j-1]*((pos[i]-j+1)>0?pos[i]-j+1:0)%p)%p;
}
LL ans=0;
if ((n+k)&1) {
printf("0\n");
return 0;
}else k=(n+k)/2;
for (int i=k;i<=n;i++){
int t=i-k,opt;
if (t&1) opt=-1;
else opt=1;
ans=(ans+opt*calc(i,k)*f
[i]%p*jc[n-i]%p)%p;
}
printf("%lld\n",(ans%p+p)%p);
}