HDU - 1402 A * B Problem Plus （FFT）

题链：https://vjudge.net/problem/HDU-1402

思路：模板题，注意前导零。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 5e4+10;
const double PI = acos(-1.0);
struct Complex
{
    double r,i;
    Complex(double _r = 0,double _i = 0)
    {
        r = _r; i = _i;
    }
    Complex operator +(const Complex &b)
    {
        return Complex(r+b.r,i+b.i);
    }
    Complex operator -(const Complex &b)
    {
        return Complex(r-b.r,i-b.i);
    }
    Complex operator *(const Complex &b)
    {
        return Complex(r*b.r-i*b.i,r*b.i+i*b.r);
    }
};
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].r /= len;
}
Complex x1[N<<2];
Complex x2[N<<2];
int n,m;
int num[N<<1]; 
char s1[N],s2[N];
int main(void){
	while(~scanf("%s%s",s1+1,s2+1)){
		n=strlen(s1+1);
		m=strlen(s2+1);
		int len,len1;
		//len1=n+m+1;
		len=1;
		while(len < 2 * n || len < 2*m ) len <<= 1;
		for(int i = 0; i < n; i++)
		   x1[i] = Complex(s1[i+1]-'0',0);
		for(int i = n; i < len; i++)
		   x1[i] = Complex(0,0);
		for(int i = 0; i < m; i++)
		   x2[i] = Complex(s2[i+1]-'0',0);
		for(int i = m; 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 <n+m;i++)
	        num[i] = (x1[i].r+0.5);
		for(int i=n+m-2;i>0;i--)
				num[i-1]+=num[i]/10,num[i]%=10;	
		int i=0;
		while(i<n+m-1&&!num[i]) i++;
		if(i==n+m-1) putchar('0');
		else{
			for(;i<n+m-1;i++)
				printf("%d",num[i]);				
		}		
		putchar('\n');
		
	}
	return 0;	
}