Bull Math
Time Limit: 1000MS | Memory Limit: 65536K |
Total Submissions: 14629 | Accepted: 7514 |
Description
Bulls are so much better at math than the cows. They can multiply huge integers together and get perfectly precise answers ... or so they say. Farmer John wonders if their answers are correct. Help him check the bulls' answers. Read in two positive integers (no more than 40 digits each) and compute their product. Output it as a normal number (with no extra leading zeros).
FJ asks that you do this yourself; don't use a special library function for the multiplication.
Input
* Lines 1..2: Each line contains a single decimal number.
Output
* Line 1: The exact product of the two input lines
Sample Input
11111111111111
1111111111
Sample Output
12345679011110987654321
Source
USACO 2004 November
題解:FFT求大數乘法。
AC代碼:
//FFT 大整數乘法
#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為i二進制反轉
{
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); //旋轉因子
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) //求卷積
{
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;
}