高精度求A*B（FFT）

A * B Problem Plus

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

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26449    Accepted Submission(s): 6917

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

 代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;

const double PI = acos(-1.0);
//复数结构体
struct complex
{
    double r,i;
    complex(double _r = 0.0,double _i = 0.0)
    {
        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);
    }
};
/*
 * 进行FFT和IFFT前的反转变换。
 * 位置i和 （i二进制反转后位置）互换
 * len必须去2的幂
 */
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]);
        //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左反转类型的+1,始终保持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^k形式，
 * on==1时是DFT，on==-1时是IDFT
 */
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;
}
const int MAXN = 200010;
complex x1[MAXN],x2[MAXN];
char str1[MAXN/2],str2[MAXN/2];
int sum[MAXN];
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    while(cin>>str1>>str2)
    {
        int len1 = strlen(str1);
        int len2 = strlen(str2);
        int len = 1;
        while(len < len1*2 || len < len2*2)len<<=1;
        for(int i = 0;i < len1;i++)
            x1[i] = complex(str1[len1-1-i]-'0',0);//这里的len1-1-i主要是为了下面的求和
        for(int i = len1;i < len;i++)
            x1[i] = complex(0,0);
        for(int i = 0;i < len2;i++)
            x2[i] = complex(str2[len2-1-i]-'0',0);
        for(int i = len2;i < len;i++)
            x2[i] = complex(0,0);
        //求DFT
        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 < len;i++)
            sum[i] = (int)(x1[i].r+0.5);
        for(int 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(int i = len;i >= 0;i--)
            printf("%c",sum[i]+'0');
        printf("
");
    }
    return 0;
}