HDU 1402 A * B Problem Plus

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

 

思路

题意很简单,就是计算两个数的乘积,像我英语这么差的人终于看懂了一道英文题目,好开心~😄

不过重要的是每一个数字的位数最多 50000 位,如果大数相乘的话也就是 50000*50000 ,一定会超时!

所以要用一种非常奇特的变换方法,那就是 FFT 傅里叶变换!

 

传送门:傅里叶变换

 

AC 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;
}