【bzoj3771】Triple FFT

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a表示一个的方案数
b表示取两个相同的
c表示取三个相同的
最终，取一个的是a
取两个的是(a*a-b)/2
取三个的是(a*a*a-3*a*b+2*z)/6

a*a*a用FFT算就可以了

乘法是序列的卷积

#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<algorithm>#include<iostream>#define maxn 200100 #define pi acos(-1)using namespace std;struct yts{double r,i;yts operator+(yts x) {yts ans;ans.r=r+x.r;ans.i=i+x.i;return ans;}yts operator-(yts x) {yts ans;ans.r=r-x.r;ans.i=i-x.i;return ans;}yts operator*(yts x) {yts ans;ans.r=r*x.r-i*x.i;ans.i=r*x.i+i*x.r;return ans;}}a[maxn],b[maxn],c[maxn],d[maxn],temp[maxn];long long ans[3][maxn];int n,m,mx,digit;int seq[maxn];void FFT(yts x[],int n,int type){if (n==1) return;for (int i=0;i<n;i+=2) temp[i>>1]=x[i],temp[i+n>>1]=x[i+1];memcpy(x,temp,sizeof(yts)*n);yts *l=x,*r=l+(n>>1);FFT(l,n>>1,type);FFT(r,n>>1,type);yts root,w;root.r=cos(2*type*pi/n),root.i=sin(2*type*pi/n);w.r=1;w.i=0;for (int i=0;i<(n>>1);i++,w=w*root)  temp[i]=l[i]+w*r[i],temp[(n>>1)+i]=l[i]-w*r[i];memcpy(x,temp,sizeof(yts)*n);}int main(){scanf("%d",&n);for (int i=1;i<=n;i++){int x;scanf("%d",&x);seq[i]=x;a[x].r++;ans[0][x]++;mx=max(mx,x);}mx*=3;for (digit=1;digit<mx;digit<<=1);FFT(a,digit,1);for (int i=0;i<=digit;i++) b[i]=a[i]*a[i],c[i]=a[i]*a[i]*a[i];FFT(b,digit,-1);FFT(c,digit,-1);for (int i=1;i<=n;i++) d[2*seq[i]].r++;FFT(d,digit,1);for (int i=0;i<=digit;i++) d[i]=d[i]*a[i];FFT(d,digit,-1);for (int i=0;i<=digit;i++){ans[1][i]=(long long)(b[i].r/digit+0.5);ans[2][i]=(long long)(c[i].r/digit+0.5)-3*(long long)(d[i].r/digit+0.5);}for (int i=1;i<=n;i++) ans[1][2*seq[i]]--,ans[2][3*seq[i]]+=2;for (int i=0;i<=digit;i++) ans[1][i]/=2,ans[2][i]/=6;for (int i=0;i<=digit;i++)  if (ans[0][i]+ans[1][i]+ans[2][i]>0) printf("%d %lld\n",i,ans[0][i]+ans[1][i]+ans[2][i]);return 0;}

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