# BZOJ3771：Triple——题解

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

（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;
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(){
for(int i=1;i<=n;i++){
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