HDU 1402 A * B Problem Plus FFT

A * B Problem Plus

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=1402

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

Hint

题意

题解：

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。

代码

#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;
}