HDU 1402:A * B Problem Plus

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 16932    Accepted Submission(s): 3558


Problem 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

 

Author

DOOM III

 

还是没有看懂这道题的代码，(；′⌒`)

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<vector>
#define ll __int64
#define pi acos(-1.0)
using namespace std;
const int MAX = 200005;
//复数结构体
struct complex
{
double r,i;
complex(double R=0,double I=0)
{
r=R;
i=I;
}
complex operator+(const complex &a)
{
return complex(r+a.r,i+a.i);
}
complex operator-(const complex &a)
{
return complex(r-a.r,i-a.i);
}
complex operator*(const complex &a)
{
return complex(r*a.r-i*a.i,r*a.i+i*a.r);
}
};
/*
*进行FFT和IFFT前的反转变换
*位置i和i的二进制反转后位置互换，(如001反转后就是100)
*len必须去2的幂
*/
void change(complex x[],int len)
{
int i,j,k;
for(i = 1, j = len/2; i <len-1; i++)
{
if (i < j) swap(x[i],x[j]);
//交换互为小标反转的元素，i<j保证交换一次
//i做正常的+1,j做反转类型的+1，始终i和j是反转的
k = len/2;
while (j >= k)
{
j -= k;
k /= 2;
}
if (j < k) j += k;
}
}
/*
*做FFT
*len必须为2^n形式，不足则补0
*on=1时是DFT，on=-1时是IDFT
*/
void fft (complex x[],int len,int on)
{
change(x,len);
for (int i=2; i<=len; i<<=1)
{
complex wn(cos(-on*2*pi/i),sin(-on*2*pi/i));
for (int j=0; j<len; j+=i)
{
complex w(1,0);
for (int k=j; k<j+i/2; k++)
{
complex u = x[k];
complex t = w*x[k+i/2];
x[k] = u+t;
x[k+i/2] = u-t;
w = w*wn;
}
}
}
if (on == -1)
for (int i=0; i<len; i++)
x[i].r /= len;
}
complex x1[MAX],x2[MAX];
char str1[MAX/2],str2[MAX/2];
ll num[MAX],sum[MAX];
int main()
{
int i,len1,len2,len;
while(scanf("%s%s",str1,str2)!=EOF)
{
len1 = strlen(str1);
len2 = strlen(str2);
len = 1;
while (len < 2*len1 || len < 2*len2) len<<=1;
for (i=0; i<len1; i++)
x1[i] = complex(str1[len1-1-i]-'0',0);
for (i=len1; i<len; i++)
x1[i] = complex(0,0);
for (i=0; i<len2; i++)
x2[i] = complex(str2[len2-1-i]-'0',0);
for (i=len2; i<len; i++)
x2[i] = complex(0,0);
fft(x1,len,1);
fft(x2,len,1);
for (i=0; i<len; i++)
x1[i] = x1[i]*x2[i];
fft(x1,len,-1);
for (i=0; i<len; i++)
sum[i] = (int)(x1[i].r+0.5);
for (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 (i=len; i>=0; i--)
printf("%c",(char)sum[i]+'0');
printf("\n");
}
return 0;
}