POJ 2389 Bull Math （FFT）

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概述

Bull Math

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 14629		Accepted: 7514

Description

Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).

FJ asks that you do this yourself; don't use a special library function for the multiplication.

Input

* Lines 1..2: Each line contains a single decimal number.

Output

* Line 1: The exact product of the two input lines

Sample Input

11111111111111
1111111111

Sample Output

12345679011110987654321

Source

USACO 2004 November

题解：FFT求大数乘法。

AC代码：

//FFT 大整数乘法
#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;
}