hdu1402 A * B Problem Plus （FFT）

最新推荐文章于 2021-02-21 10:12:55 发布

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最新推荐文章于 2021-02-21 10:12:55 发布

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原文链接：http://www.cnblogs.com/q934098774/p/5388699.html

本文介绍了一种使用快速傅里叶变换（FFT）来高效地计算多项式乘法的方法，通过将多项式转化为系数形式，利用FFT进行运算，最终实现多项式的快速乘法。

思路：考虑变成系数的形式，显然就是两个的多项式乘法，然后转化成FFT，直接莽一波就完了。

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>

using namespace std;


const int N = 500005;
const double pi = acos(-1.0);

char s1[N],s2[N];
int len,res[N];

struct Complex
{
    double r,i;
    Complex(double r=0,double i=0):r(r),i(i) {};
    Complex operator+(const Complex &rhs)
    {
        return Complex(r + rhs.r,i + rhs.i);
    }
    Complex operator-(const Complex &rhs)
    {
        return Complex(r - rhs.r,i - rhs.i);
    }
    Complex operator*(const Complex &rhs)
    {
        return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);
    }
} va[N],vb[N];

void rader(Complex F[],int len) //len = 2^M,reverse F[i] with  F[j] j为i二进制反转
{
    int j = len >> 1;
    for(int i = 1;i < len - 1;++i)
    {
        if(i < j) swap(F[i],F[j]);  // reverse
        int k = len>>1;
        while(j>=k)
        {
            j -= k;
            k >>= 1;
        }
        if(j < k) j += k;
    }
}

void FFT(Complex F[],int len,int t)
{
    rader(F,len);
    for(int h=2;h<=len;h<<=1)
    {
        Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h));
        for(int j=0;j<len;j+=h)
        {
            Complex E(1,0); //旋转因子
            for(int k=j;k<j+h/2;++k)
            {
                Complex u = F[k];
                Complex v = E*F[k+h/2];
                F[k] = u+v;
                F[k+h/2] = u-v;
                E=E*wn;
            }
        }
    }
    if(t==-1)   //IDFT
        for(int i=0;i<len;++i)
            F[i].r/=len;
}

void Conv(Complex a[],Complex b[],int len) //求卷积
{
    FFT(a,len,1);
    FFT(b,len,1);
    for(int i=0;i<len;++i) a[i] = a[i]*b[i];
    FFT(a,len,-1);
}

void init(char *s1,char *s2)
{
    int n1 = strlen(s1),n2 = strlen(s2);
    len = 1;
    while(len < 2*n1 || len < 2*n2) len <<= 1;
    int i;
    for(i=0;i<n1;++i)
    {
        va[i].r = s1[n1-i-1]-'0';
        va[i].i = 0;
    }
    while(i<len)
    {
        va[i].r = va[i].i = 0;
        ++i;
    }
    for(i=0;i<n2;++i)
    {
        vb[i].r = s2[n2-i-1]-'0';
        vb[i].i = 0;
    }
    while(i<len)
    {
        vb[i].r = vb[i].i = 0;
        ++i;
    }
}

void gao()
{
    Conv(va,vb,len);
    memset(res,0,sizeof res);
    for(int i=0;i<len;++i)
    {
        res[i]=va[i].r + 0.5;
    }
    for(int i=0;i<len;++i)
    {
        res[i+1]+=res[i]/10;
        res[i]%=10;
    }
    int high = 0;
    for(int i=len-1;i>=0;--i)
    {
        if(res[i])
        {
            high = i;
            break;
        }
    }
    for(int i=high;i>=0;--i) putchar('0'+res[i]);
    puts("");
}


int main()
{
    while(scanf("%s %s",s1,s2)==2)
    {
        init(s1,s2);
        gao();
    }
    return 0;
}

Description

Calculate A * B.

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output

For each case, output A * B in one line.

Sample Input