bzoj2194: 快速傅立叶之二

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=400005;
const double PI=acos(-1);
struct node{
    double real,imag;
    void clear(){real=imag=0;}
    node operator +(const node &x){return (node){real+x.real,imag+x.imag};}
    node operator -(const node &x){return (node){real-x.real,imag-x.imag};}
    node operator *(const node &x){return (node){real*x.real-imag*x.imag,real*x.imag+imag*x.real};}
}a[maxn],b[maxn],c[maxn],t1,t2,w,wn;
int m,n,len,rev[maxn],ans[maxn];
void read(int &x){
    x=0; int f=1; char ch;
    for (ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
    for (;isdigit(ch);ch=getchar()) x=x*10+ch-'0'; x*=f;
}
int Rev(int x){
    int temp=0;
    for (int i=1;i<=len;i++) temp<<=1,temp+=(x&1),x>>=1;
    return temp;
}
void FFT(node *a,int op){
    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=(node){cos(2*op*PI/s),sin(2*op*PI/s)};
        for (int i=0;i<n;i+=s){
            w=(node){1,0};
            for (int j=i;j<i+s/2;j++,w=w*wn){
                t1=a[j],t2=w*a[j+s/2];
                a[j]=t1+t2,a[j+s/2]=t1-t2;
            }
        }
    }
}
int main(){
    read(m); n=1,len=0;
    while (n<(m<<1)) n<<=1,len++;
    for (int i=0;i<n;i++) rev[i]=Rev(i);
    for (int x,i=0;i<m;i++) read(x),a[i].real=x,read(x),b[m-1-i].real=x;
    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)round(c[i].real/n);
    for (int i=m-1;i<2*m-1;i++) printf("%d
",ans[i]);
    return 0;
}

题目链接：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。

做法：考虑把b数组翻转，Ck的计算就成为了裸的卷积，对于这种题目，翻转是个重要的手段。