BZOJ2194: 快速傅立叶之二

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2194

题目大意：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。

题解：这就是所谓的卷积，找个时间一定要好好看看，上FFT咯

代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<cmath>
#define maxn 270005
#define pi acos(-1)
using namespace std;
int n,m;
int ans[maxn],rev[maxn];
int l;
struct F{
    double rea,ima;
    F operator +(const F &x){return (F){rea+x.rea,ima+x.ima};}
    F operator -(const F &x){return (F){rea-x.rea,ima-x.ima};}
    F operator *(const F &x){return (F){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}
}a[maxn],b[maxn],c[maxn],w,wn,t1,t2;
int read()
{
    int x=0; char ch; bool bo=0;
    while (ch=getchar(),ch<'0'||ch>'9') if (ch=='-') bo=1;
    while (x=x*10+ch-'0',ch=getchar(),ch>='0'&&ch<='9');
    if (bo) return -x; return x;
}
void fft(F *a,int type)
{
    for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int s=2; s<=n; s<<=1)
    {
        wn=(F){cos(type*2*pi/s),sin(type*2*pi/s)};
        for (int i=0; i<n; i+=s)
        {
            w=(F){1,0};
            for (int j=i; j<(i+(s>>1)); j++,w=w*wn)
            {
                t1=a[j],t2=a[j+(s>>1)]*w;
                a[j]=t1+t2,a[j+(s>>1)]=t1-t2;
            }
        }
    }
}
int re(int x)
{
    int t=0;
    for (int i=0; i<l; i++) t<<=1,t|=x&1,x>>=1;
    return t;
}
void init()
{
    m=read();
    for (n=1; n<2*m; n<<=1) l++;
    for (int i=0; i<m; i++) 
    {
        a[i].rea=read(),b[m-i-1].rea=read();
    }
    for (int i=0; i<n; i++) rev[i]=re(i);
    fft(a,1); fft(b,1);
    for (int i=0; i<n; i++) c[i]=a[i]*b[i];
    fft(c,-1);
    for (int i=0; i<n; i++) ans[i]=(int)(c[i].rea/n+0.5);
    for (int i=m-1; i<2*m-1; i++) printf("%d
",ans[i]);
    
}
int main()
{
    init();
}