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
,s2
;
int len,res
;

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
,vb
;

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;
}