[BZOJ 2194] Fast Fourier second, bzoj2194 Fourier

[BZOJ 2194] Fast Fourier second, bzoj2194 Fourier

2194: Fast Fourier second

Time Limit: 10 Sec Memory Limit: 259 MB
Submit: 430 Solved: 240
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Description

Calculate C [k] = sigma (a [I] * B [I-k]), where k <= I <n and n <= 10 ^ 5. All elements in a and B are non-negative integers less than or equal to 100.

Input

The first line is an integer N, And the next N rows are I + 2 .. I + N-1 line, two numbers in each row, in turn represents a [I], B [I] (0 <= I <N ).

Output

Output N rows, each row has an integer, and the I rows Output C [I-1].

Sample Input

5

3 1

2 4

1 1

2 4

1 4

Sample Output

24

12

10

6

1

FFT template question.

For more information about FFT, see [BZOJ 2179]

FFT calculates the product of two polynomials whose subscript is a fixed value, and the difference in this question is a fixed value. We only need to reverse the order of the [] array to the sum value!

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <complex>#define N 500005#define pi acos(-1)#define LL long longusing namespace std;int n;complex<double> a[N],b[N],p[N];void read(int &tmp){    tmp=0;    char ch=getchar();    int fu=1;    for (;ch<'0'||ch>'9';ch=getchar())        if (ch=='-') fu=-1;    for (;ch>='0'&&ch<='9';ch=getchar())        tmp=tmp*10+ch-'0';    tmp*=fu;}void FFT(complex<double> x[],int n,int p){    for (int i=0,t=0;i<n;i++)    {        if (i>t) swap(x[i],x[t]);        for (int j=n>>1;(t^=j)<j;j>>=1);    }    for (int m=2;m<=n;m<<=1)    {        complex<double> wn(cos(p*2*pi/m),sin(p*2*pi/m));        for (int i=0;i<n;i+=m)        {            complex<double> w(1,0),u;            int k=m>>1;            for (int j=0;j<k;j++,w*=wn)            {                u=x[i+j+k]*w;                x[i+j+k]=x[i+j]-u;                x[i+j]=x[i+j]+u;            }        }    }}int main(){    scanf("%d",&n);    for (int i=0;i<n;i++)    {        int x;        read(x);        a[n-i-1]=x;        read(x);        b[i]=x;    }    int nn=n;    for (int j=n,i=1;(i>>1)<j;i<<=1)        n=i;    cout<<n<<endl;    FFT(a,n,1),FFT(b,n,1);    for (int i=0;i<n;i++)        p[i]=a[i]*b[i];    FFT(p,n,-1);    for (int i=nn-1;i>=0;i--)        printf("%lld\n",(LL)(p[i].real()/n+0.5));    return 0;}