Description

请计算C[k]=sigma(a[i]*b[i-k]) 其中 k < = i < n ，并且有 n < = 10 ^ 5。 a,b中的元素均为小于等于100的非负整数。

Input

第一行一个整数N,接下来N行，第i+2..i+N-1行，每行两个数，依次表示a[i],b[i] (0 < = i < N)。

Output

输出N行，每行一个整数，第i行输出C[i-1]。

Sample Input

5
3 1
2 4
1 1
2 4
1 4

Sample Output

24
12
10
6
1

Solution

看上去是个FFT的模板题，实际上它就是的
将b数组翻转之后，c数组就可以用FFT求了
手写c数组原来一些位置的式子，然后会发现它们在新的c数组的位置的规律
输出就好了

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
const db Pi=acos(-1);
const int MAXN=1<<19;
int n,m,cnt,rev[MAXN],sn;
struct Complex{db real,imag;inline Complex operator + (const Complex &A) const {return (Complex){real+A.real,imag+A.imag};};inline Complex operator - (const Complex &A) const {return (Complex){real-A.real,imag-A.imag};};inline Complex operator * (const Complex &A) const {return (Complex){real*A.real-imag*A.imag,imag*A.real+real*A.imag};};
};
Complex a[MAXN],b[MAXN];
template<typename T> inline void read(T &x)
{T data=0,w=1;char ch=0;while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();if(ch=='-')w=-1,ch=getchar();while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+'0');if(ch!='\0')putchar(ch);
}
template<typename T> inline void chkmin(T &x,T y){x=(y<x?y:x);}
template<typename T> inline void chkmax(T &x,T y){x=(y>x?y:x);}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void FFT(Complex *A,int tp)
{for(register int i=0;i<n;++i)if(i<rev[i])std::swap(A[i],A[rev[i]]);for(register int l=2;l<=n;l<<=1){Complex wn=(Complex){cos(2*Pi/l),sin(tp*2*Pi/l)};for(register int i=0;i<n;i+=l){Complex w=(Complex){1,0};for(register int j=0;j<(l>>1);++j){Complex A1=A[i+j],A2=A[i+j+(l>>1)]*w;A[i+j]=A1+A2,A[i+j+(l>>1)]=A1-A2;w=w*wn;}}}
}
int main()
{read(n);m=n+n-1;sn=n;for(register int i=0;i<n;++i)scanf("%lf%lf",&a[i].real,&b[i].real);std::reverse(b,b+n);for(n=1;n<m;n<<=1)++cnt;for(register int i=0;i<n;++i)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));FFT(a,1);FFT(b,1);for(register int i=0;i<n;++i)a[i]=a[i]*b[i];FFT(a,-1);for(register int i=sn-1;i<=sn+sn-2;++i)write((int)(a[i].real/n+0.5),'\n');return 0;
}