hdu 1402 FFT（模板）

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16111    Accepted Submission(s): 3261

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

 

Recommend

题意：求高精度a*b                                  --代码参考kuangbin大神

思路：

通过FFT我们可以快速求出多项式的卷积，从而解决数相乘。

求卷积大致如下图，至于FFT具体原理看不太懂- -

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;
const ld eps=1e-10;
const int inf = 0x3f3f3f;
const int MOD = 1e9+7;

const double PI = acos(-1.0);

struct Complex
{
    double x,y;
    Complex(double _x = 0.0,double _y = 0.0)
    {
        x = _x;
        y = _y;
    }
    Complex operator-(const Complex &b)const
    {
        return Complex(x-b.x,y-b.y);
    }
    Complex operator+(const Complex &b)const
    {
        return Complex(x+b.x,y+b.y);
    }
    Complex operator*(const Complex &b)const
    {
        return Complex(x*b.x-y*b.y,x*b.y+y*b.x);
    }
};

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]);
        k = len/2;
        while(j >= k)
        {
            j-=k;
            k/=2;
        }
        if(j < k) j+=k;
    }
}

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].x /= len;
    }
}

const int maxn = 200100;
Complex x1[maxn],x2[maxn];
char str1[maxn],str2[maxn];
int sum[maxn];

int main()
{
    while(scanf("%s%s",str1,str2) != EOF)
    {
        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-i-1]-'0',0);
        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);

        fft(x1,len,1);
        fft(x2,len,1);
        for(int i = 0; i < len; i++)
        {
            x1[i] =x1[i]*x2[i];
            //cout << x1[i].x << " "<< x1[i].y <<endl;
        }
        fft(x1,len,-1);
        for(int i = 0;i < len;i++){
            sum[i] = (int)(x1[i].x+0.5);
            //cout << sum[i] << endl;
        }

        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("\n");
    }
    return 0;
}