BZOJ3771:Triple——题解

https://www.lydsy.com/JudgeOnline/problem.php?id=3771

大意:给n把不同价值的斧子,从中选一把/两把/三把,所构成的每种价值和的可能情况有多少。

生成函数,指数为价值,系数即为可能情况数。

但是直接FFT乘会有两把/三把斧子拿的同一个的情况。

于是我们多存两个数组,分别记录两把/三把同时拿一把的生成函数,之后就容斥一下就行啦!

(注意拿的顺序不同也算是同一种情况,不要忘记除下去)

#include<cstdio>
#include<cctype>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
typedef long double dl;
const dl pi=acos(-1.0);
const int N=3e5+5;
inline int read(){
    int X=0,w=0;char ch=0;
    while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
    while(isdigit(ch))X=(X<<3)+(X<<1)+(ch^48),ch=getchar();
    return w?-X:X;
}
struct complex{//定义复数 
    dl x,y;
    complex(dl xx=0.0,dl yy=0.0){
        x=xx;y=yy;
    }
    complex operator +(const complex &b)const{
        return complex(x+b.x,y+b.y);
    }
    complex operator -(const complex &b)const{
        return complex(x-b.x,y-b.y);
    }
    complex operator *(const complex &b)const{
        return complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
};
void FFT(complex a[],int n,int on){
    for(int i=1,j=n>>1;i<n-1;i++){
        if(i<j)swap(a[i],a[j]);
        int k=n>>1;
        while(j>=k){j-=k;k>>=1;}
        if(j<k)j+=k;
    }
    for(int i=2;i<=n;i<<=1){
        complex res(cos(-on*2*pi/i),sin(-on*2*pi/i));
        for(int j=0;j<n;j+=i){
            complex w(1,0);
            for(int k=j;k<j+i/2;k++){
                complex u=a[k],t=w*a[k+i/2];
                a[k]=u+t;
                a[k+i/2]=u-t;
                w=w*res;
            }
        }
    }
    if(on==-1)
        for(int i=0;i<n;i++)a[i].x/=n;
}
int n,m;
complex a[N],b[N],c[N],d[N];
int main(){
    n=read();
    for(int i=1;i<=n;i++){
    int w=read();m=max(m,w);
    a[w].x=1;
    b[w*2].x=1;
    c[w*3].x=1;
    }
    m=m*3;
    int nn=1;
    while(nn<m)nn<<=1;
    FFT(a,nn,1);FFT(b,nn,1);FFT(c,nn,1);
    for(int i=0;i<nn;i++){
    complex t1(1.0/2.0,0);
    complex t2(3.0,0);
    complex t3(2.0,0);
    complex t4(1.0/6.0,0);
    d[i]=d[i]+a[i];
    d[i]=d[i]+(a[i]*a[i]-b[i])*t1;
    d[i]=d[i]+(a[i]*a[i]*a[i]-t2*a[i]*b[i]+t3*c[i])*t4;
    }
    FFT(d,nn,-1);
    for(int i=0;i<m;i++){
    int w=d[i].x+0.5;
    if(w)printf("%d %d\n",i,w);
    }
    return 0;
}

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+本文作者:luyouqi233。               +

+欢迎访问我的博客:http://www.cnblogs.com/luyouqi233/ +

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posted @ 2018-05-07 15:34  luyouqi233  阅读(177)  评论(0编辑  收藏  举报