BZOJ 3771: Triple

Description

问所有三/二/一元组可能形成的组合.

Sol

FFT.

利用生成函数直接FFT一下,然后就是计算,计算的时候简单的容斥一下.

任意三个-3*两个相同的+2*全部相同的+任意两个-两个相同的+任意一个.

Code

/**************************************************************
    Problem: 3771
    User: BeiYu
    Language: C++
    Result: Accepted
    Time:1760 ms
    Memory:26684 kb
****************************************************************/
 
#include <bits/stdc++.h>
using namespace std;
 
#define mpr make_pair
#define rr first
#define ii second
typedef pair< double,double > Complex;
const int N = 5e5+50;
const int M = 40000;
const double Pi = M_PI;
 
Complex operator + (const Complex &a,const Complex &b) {
    return mpr(a.rr+b.rr,a.ii+b.ii);
}
Complex operator - (const Complex &a,const Complex &b) {
    return mpr(a.rr-b.rr,a.ii-b.ii);
}
Complex operator * (const Complex &a,const Complex &b) {
    return mpr(a.rr*b.rr-a.ii*b.ii,a.rr*b.ii+a.ii*b.rr);
}
Complex operator * (const Complex &a,const int &b) {
    return mpr(a.rr*b,a.ii*b);
}
Complex operator / (const Complex &a,const double &b) {
    return mpr(a.rr/b,a.ii/b);
}
int n;
Complex a[N],b[N],c[N];
int ans[N];
 
void Rev(Complex a[]) {
    for(int i=0,j=0;i<n;i++) {
        if(i>j) swap(a[i],a[j]);
        for(int k=n>>1;(j^=k)<k;k>>=1);
    }
}
void DFT(Complex a[],int r) {
    Rev(a);
    for(int i=2;i<=n;i<<=1) {
        Complex wi=mpr(cos(2.0*Pi/i),r*sin(2.0*Pi/i));
        for(int k=0;k<n;k+=i) {
            Complex w=mpr(1.0,0.0);
            for(int j=k;j<k+i/2;j++) {
                Complex t1=a[j],t2=w*a[j+i/2];
                a[j]=t1+t2,a[j+i/2]=t1-t2;
                w=w*wi;
            }
        }
    }if(r==-1) for(int i=0;i<n;i++) a[i].rr/=n;
}
void FFT(Complex a[],Complex b[],Complex c[]) {
    DFT(a,1),DFT(b,1);
    for(int i=0;i<n;i++) c[i]=a[i]*b[i];
    DFT(c,-1);
}
void init(int x) {
    for(n=1;n<x;n<<=1);
}
void Print(Complex a[]) {
    for(int i=0;i<n;i++) if((int)(a[i].rr+0.5)>=1) printf("%d %d
",i,int(a[i].rr+0.5));
}
int main() {
    int l;
    cin>>l;
    for(int i=1,x;i<=l;i++) cin>>x,a[x].rr+=1.0,b[x*2].rr+=1.0,c[x*3].rr+=1.0;
    init(M*3);
     
    DFT(a,1),DFT(b,1),DFT(c,1);
    for(int i=0;i<n;i++) a[i]=(a[i]*a[i]*a[i]-a[i]*b[i]*3+c[i]*2)/6.0+(a[i]*a[i]-b[i])/2.0+a[i];
    DFT(a,-1);
     
    Print(a);
    return 0;
}