标签:problem space type amp namespace 组合 result cond cpp
问所有三/二/一元组可能形成的组合.
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.
/************************************************************** 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\n",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; }
标签:problem space type amp namespace 组合 result cond cpp
原文地址:http://www.cnblogs.com/beiyuoi/p/6254870.html