BZOJ 2179FFT快速傅立叶

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=2179

题目大意：给出两个n位10进制整数x和y，你需要计算x*y。

题解：FFT，不会的可以膜拜陈老师（非clj）QQ：297086016

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define inf 1<<30
#define maxm 270005
#define pi acos(-1)
using namespace std;
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[maxm],b[maxm],c[maxm],w,wn,t1,t2;
int n,m,l,ans[maxm],rev[maxm];
void read(F *a)
{
    char ch;
    while (ch=getchar(),ch<'0'||ch>'9');
    for (int i=m-1; ch>='0'&&ch<='9';ch=getchar(),i--) a[i].rea=ch-'0';
}
int re(int v)
{
    int t=0;
    for (int i=0; i<l; i++) t<<=1,t|=v&1,v>>=1;
    return t;
}
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=w*a[j+(s>>1)];
                a[j]=t1+t2,a[j+(s>>1)]=t1-t2;
            }
        }
    }
}
int main()
{
    scanf("%d\n",&m);
    for (n=1; n<(m<<1); n<<=1) l++;
    for (int i=0; i<n; i++) rev[i]=re(i);
    read(a); read(b);
    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=0;i<n;i++) ans[i+1]+=ans[i]/10,ans[i]=ans[i]%10;
    int pps=n-1;while(ans[pps]==0&&pps)pps--;
    for(;pps>=0;pps--)printf("%d",ans[pps]);printf("\n");
    return 0;
}