BZOJ2179:FFT快速傅立叶(FFT)

Description

给出两个n位10进制整数x和y，你需要计算x*y。

Input

第一行一个正整数n。 第二行描述一个位数为n的正整数x。 第三行描述一个位数为n的正整数y。

Output

输出一行，即x*y的结果。

Sample Input

1
3
4

Sample Output

12
数据范围：
n<=60000

Solution

FFT模板题，做的时候注意处理一下进位和前导零就好

Code

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#define N (240000+100)
using namespace std;
double pi=acos(-1.0);
struct complex
{
    double x,y;
    complex (double xx=0,double yy=0)
    {
        x=xx; y=yy;
    }
}a[N],b[N];
int n=-1,m=-1,t,fn,l,r[N];
char ch;

complex operator + (complex a,complex b){return complex(a.x+b.x,a.y+b.y);}
complex operator - (complex a,complex b){return complex(a.x-b.x,a.y-b.y);}
complex operator * (complex a,complex b){return complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
complex operator / (complex a,double b){return complex(a.x/b,a.y/b);}

void FFT(int n,complex *a,int opt)
{
    for (int i=0; i<n; ++i)
        if (i<r[i])
            swap(a[i],a[r[i]]);
    for (int k=1; k<n; k<<=1)
    {
        complex wn=complex(cos(pi/k),opt*sin(pi/k));
        for (int i=0; i<n; i+=(k<<1))
        {
            complex w=complex(1,0);
            for (int j=0; j<k; ++j,w=w*wn)
            {
                complex x=a[i+j], y=w*a[i+j+k];
                a[i+j]=x+y; a[i+j+k]=x-y;
            }
        }
    }
    if (opt==-1) for (int i=0; i<n; ++i) a[i]=a[i]/n;
}

int main()
{
    scanf("%d",&t); t--;
    for (int i=0; i<=t; ++i)
    {
        ch=getchar();
        while (ch<'0' || ch>'9') ch=getchar();
        if (n==-1 && ch=='0') continue;
        a[++n].x=ch-'0';
    }
    for (int i=0; i<=t; ++i)
    {
        ch=getchar();
        while (ch<'0' || ch>'9') ch=getchar();
        if (m==-1 && ch=='0') continue;
        b[++m].x=ch-'0';
    }
    if (m==-1) m=0;
    if (n==-1) n=0;
    fn=1;
    while (fn<=n+m) fn<<=1, l++;
    for (int i=0;i<fn;++i)
        r[i]=(r[i>>1]>>1) | ((i&1)<<(l-1));
    
    FFT(fn,a,1); FFT(fn,b,1);
    for (int i=0;i<=fn;++i)
        a[i]=a[i]*b[i];
    FFT(fn,a,-1);
    for (int i=n+m;i>=1;--i)
    {
        a[i-1].x+=(int)(a[i].x+0.5)/10;
        a[i].x=(int)(a[i].x+0.5)%10;
    }
    int p=0;
    while ((int)(a[p].x+0.5)==0 && p<n+m) p++; 
    for (int i=p;i<=n+m;++i)
        printf("%d",(int)(a[i].x+0.5));
}