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先将b数组从小到大排序

设选中了xa > b,由总对数为n,由x-(n-x)=k,可以知道x=\frac{n+k}2

我们设f(i,j)为前ia中选中了ja > b的方案数

那么f(i,j)=f(i-1,j)+f(i-1,j-1)\times(l_i-j+1)

(l_i表示b中小于a_i的最后一个位置)

但是还有剩下的n-x

我们可以设g_i表示a>b对数\ge i的方案数

那么g_i = f(n,i) \times (n-i)!(相当于剩下的随便排列组合)

我们设f_i表示a>b对数恰好为i对的方案数

那么g_k = \sum_{i=k}^n f_i \times {i\choose k} (相当于从恰好i个方案中选k对出来)

经过二项式反演可以知道:

f(k)=\sum_{i=k}^n(-1)^{i-k}{i\choose k}g(i)

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;rg char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;rg char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
#define N 4011
const int mod = 1000000009;
int n,k,f[N][N],a[N],b[N],l[N],fac[N],inv[N],g[N];
il int pw(int x,int p){
    int ans=1;
    while(p){
        if(p&1)ans=1ll*ans*x%mod;
        p>>=1;x=1ll*x*x%mod;
    }
    return ans;
}
int C(int a,int b){
    return 1ll*fac[a]*inv[b]%mod*inv[a-b]%mod;
}
signed main(){
    in(n,k);
    if((n+k)&1){puts("0");return 0;};
    k=(n+k)>>1;
    Fur(i,1,n)in(a[i]);
    Fur(i,1,n)in(b[i]);
    fac[0]=inv[0]=1;
    Fur(i,1,n)fac[i]=1ll*fac[i-1]*i%mod;
    inv[n]=pw(fac[n],mod-2);
    Fdr(i,n-1,0)inv[i]=1ll*inv[i+1]*(i+1)%mod;
    sort(a+1,a+n+1);
    sort(b+1,b+n+1);
    int p=0;
    Fur(i,1,n){
        while(p<n&&b[p+1]<a[i])++p;
        l[i]=p;
    }
    f[0][0]=1;
    Fur(i,1,n){
        f[i][0]=1;
        Fur(j,1,i)
            f[i][j]=(f[i-1][j]+1ll*f[i-1][j-1]*MAX(0,l[i]-j+1)%mod)%mod;
    }
    Fur(i,0,n)g[i]=1ll*f[n][i]*fac[n-i]%mod;
    int ans=0;
    Fur(i,k,n)
        if((i-k)&1)ans=(ans-1ll*C(i,k)*g[i]%mod+mod)%mod;
        else ans=(ans+1ll*C(i,k)*g[i]%mod)%mod;

    printf("%d\n",ans);
}
LG 4859 已经没有什么好害怕的了
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