A * B Problem Plus HDU - 1402 （FFT）

Posted 2021-02-06 smallhester

A * B Problem Plus HDU - 1402 （FFT）

Calculate A * B. 

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

Note: the length of each integer will not exceed 50000. 
OutputFor each case, output A * B in one line. 
Sample Input

1
2
1000
2

Sample Output

2
2000


题意：求A*B，A和B的长度都小于50000
题解：FFT的板子题，但FFT还不会，之后再贴一些想法

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<map>
#include<cstdlib>
#include<vector>
#include<string>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn =  2e5+10;
const int mod = 1e9+7;


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