bzoj2194: 快速傅立叶之二

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

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。