BZOJ 3527 力（FFT）

Description

给出一个长度为n的序列q0,...,n−1，定义Fj=∑i<jqiqj(i−j)2−∑i>jqiqj(i−j)2，求Ei=Fiqi

Input

第一行一个整数n表示序列长度，之后输入n个实数q0,...,n−1(n≤105,0<qi<109)

Output

输出n个实数E0,...,n−1，结果与标准答案误差不超过10−2

Sample Input

5

4006373.885184

15375036.435759

1717456.469144

8514941.004912

1410681.345880

Sample Output

-16838672.693

3439.793

7509018.566

4595686.886

10903040.872

Solution

Ei=∑j=0i−1qj1(i−j)2−∑j=i+1n−1qj1(i−j)2=∑j=0i−1qj1(i−j)2−∑j=1n−i−1qn−j1(n−i−j)2

令A[i]=qi,B[i]=qn−i,0≤i<n,A[n]=B[n]=0,C[i]=1i2,1≤i≤n,C[0]=0

令X[i]=∑j=0iA[j]C[i−j],Y[i]=∑j=0iB[j]C[i−j],0≤i≤n

做两遍FFT得到X和Y，进而由Ei=X[i]−Y[n−i]求得E

Code

#include<cstdio>#include<cmath>#include<algorithm> using namespace std;#define maxn 100005#define maxfft 131072+5const double pi=acos(-1.0);struct cp{    double a,b;    cp operator +(const cp &o)const {return (cp){a+o.a,b+o.b};}    cp operator -(const cp &o)const {return (cp){a-o.a,b-o.b};}    cp operator *(const cp &o)const {return (cp){a*o.a-b*o.b,b*o.a+a*o.b};}    cp operator *(const double &o)const {return (cp){a*o,b*o};}    cp operator !() const{return (cp){a,-b};}}w[maxfft];int pos[maxfft];void fft_init(int len){    int j=0;    while((1<<j)<len)j++;    j--;    for(int i=0;i<len;i++)        pos[i]=pos[i>>1]>>1|((i&1)<<j);}void fft(cp *x,int len,int sta){    for(int i=0;i<len;i++)        if(i<pos[i])swap(x[i],x[pos[i]]);    w[0]=(cp){1,0};    for(unsigned i=2;i<=len;i<<=1)    {        cp g=(cp){cos(2*pi/i),sin(2*pi/i)*sta};        for(int j=i>>1;j>=0;j-=2)w[j]=w[j>>1];        for(int j=1;j<i>>1;j+=2)w[j]=w[j-1]*g;        for(int j=0;j<len;j+=i)        {            cp *a=x+j,*b=a+(i>>1);            for(int l=0;l<i>>1;l++)            {                cp o=b[l]*w[l];                b[l]=a[l]-o;                a[l]=a[l]+o;            }        }    }    if(sta==-1)for(int i=0;i<len;i++)x[i].a/=len,x[i].b/=len;}cp x[maxfft],y[maxfft],z[maxfft];void FFT(double *a,double *b,int n,int m,double *c){    int len=1;    while(len<(n+m)>>1)len<<=1;    fft_init(len);    for(int i=n/2;i<len;i++)x[i].a=x[i].b=0;    for(int i=m/2;i<len;i++)y[i].a=y[i].b=0;    for(int i=0;i<n;i++)(i&1?x[i>>1].b:x[i>>1].a)=a[i];    for(int i=0;i<m;i++)(i&1?y[i>>1].b:y[i>>1].a)=b[i];    fft(x,len,1),fft(y,len,1);    for(int i=0;i<len/2;i++)    {        int j=len-1&len-i;        z[i]=x[i]*y[i]-(x[i]-!x[j])*(y[i]-!y[j])*(w[i]+(cp){1,0})*0.25;    }    for(int i=len/2;i<len;i++)    {        int j=len-1&len-i;        z[i]=x[i]*y[i]-(x[i]-!x[j])*(y[i]-!y[j])*((cp){1,0}-w[i^len>>1])*0.25;    }    fft(z,len,-1);    for(int i=0;i<n;i++)        if(i&1)c[i]=z[i>>1].b;        else c[i]=z[i>>1].a;}int n;double q[maxn],A[maxn],B[maxn],ans[maxn];int main(){    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)scanf("%lf",&q[i]);        for(int i=0;i<n;i++)A[i]=q[i];        A[n]=0;        B[0]=0;        for(int i=1;i<=n;i++)B[i]=1.0/i/i;        FFT(A,B,n+1,n+1,A);        for(int i=0;i<n;i++)ans[i]=A[i];        for(int i=0;i<n;i++)A[i]=q[n-i];        A[n]=0;        B[0]=0;        for(int i=1;i<=n;i++)B[i]=1.0/i/i;        FFT(A,B,n+1,n+1,A);        for(int i=0;i<n;i++)ans[i]-=A[n-i];        for(int i=0;i<n;i++)            printf("%.3f\n",ans[i]);    }    return 0;}