FFT 快速傅立叶变换 dhu1402 A*B

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14007    Accepted Submission(s): 2538

Problem 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

    
    
    
    

     
     
     
     1
2
1000
2
    
    
    
    

 

Sample Output

    
    
    
    

     
     
     
     2
2000
    
    
    
    

 

Author

DOOM III

 

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FFT模板题：求A*B，用直接相乘TLE(n^2)，可以用FFT(nlogn)优化求解。

FFT原理我也理解了好久，这里FFT与多项式相乘的关系还是比较容易理解的。

如图：

附上一个连接： FFT与多项式相乘

#include <iostream>
#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <string.h>

#define     LL                                 long long
#define     scan(a)                         scanf("%d",&a)
#define     REP(i,a,b)                     for(int i=a;i<b;++i)
#define     mset(a,b)                     memset(a,b,sizeof a)
//#define     maxn                            1e6+10

using namespace std;

const double PI = acos(-1.0);


const LL mod = 998244353;
const int maxn = 2e5+10;


struct Complex
{
    double x,y;

    Complex(double _x = 0.0,double _y=0.0)
    {
        x=_x;
        y=_y;
    }
    Complex operator -(const Complex &b) const
    {
        return Complex(x-b.x,y-b.y);
    }
    Complex operator +(const Complex &b)const
    {
        return Complex(x+b.x,y+b.y);
    }
    Complex operator *(const Complex &b)const
    {
        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
};


void change(Complex y[],int len)
{
    int i,j,k;
    for(int i=1,j=len/2;i<len-1;++i)
    {
        if(i<j) swap(y[i],y[j]);
        k=len/2;
        while(j>=k)
        {
            j-=k;
            k/=2;

        }
        if(j<k) j+=k;
    }
}

void fft(Complex y[],int len,int on)
{
    change(y,len);
    for(int h=2;h<=len;h<<=1)
    {
        Complex wn(cos(-on*2*PI/h),sin(-on*2*PI/h));
        for(int j=0;j<len;j+=h)
        {
            Complex w(1,0);
            for(int k=j;k<j+h/2;k++)
            {
                Complex u=y[k];
                Complex t=w*y[k+h/2];
                y[k]=u+t;
                y[k+h/2]=u-t;
                w=w*wn;
            }
        }
    }
    if(on==-1)
        for(int i=0;i<len;++i)
            y[i].x /=len;
}

Complex x1[maxn],x2[maxn];
char str1[maxn],str2[maxn];
int sum[maxn];

int main()
{
    while(~scanf("%s%s",&str1,&str2))
    {
        int len1=strlen(str1);
        int len2=strlen(str2);
        int len=1;
        while(len<len1*2 || len<len2*2) len<<=1;
        REP(i,0,len1)
            x1[i]=Complex(str1[len1-1-i]-'0',0);
        REP(i,len1,len)
            x1[i]=Complex(0,0);
        REP(i,0,len2)
            x2[i]=Complex(str2[len2-i-1]-'0',0);
        REP(i,len2,len)
            x2[i]=Complex(0,0);
        fft(x1,len,1);
        fft(x2,len,1);
        REP(i,0,len)
            x1[i]=x1[i]*x2[i];
        fft(x1,len,-1);
        mset(sum,0);
        REP(i,0,len)
        {
            sum[i]+=int(x1[i].x+0.5);
        }
        REP(i,0,len)
        {
            sum[i+1]+=sum[i]/10;
            sum[i]%=10;
        }
        len=len1+len2-1;
        while(sum[len] <= 0 && len>0)
            len--;
        REP(i,0,len+1)
            printf("%c",sum[len-i]+'0');
        puts("");
    }
}