HDU1402（fft）

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22082    Accepted Submission(s): 5511

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

 

//2017-09-07
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <complex>
#include <cmath>
#define Complex complex<double>

using namespace std;

const double PI = acos(-1.0);
const int N = 110000;

void fft(Complex y[], int n, int op){
    for(int i = 1, j = n/2; i < n-1; i++){
        if(i<j)swap(y[i], y[j]);
        int k = n/2;
        while(j >= k){
            j -= k;
            k /= 2;
        }
        if(j<k)j += k;
    }
    for(int h = 2; h <= n; h <<= 1){
        Complex wn(cos(-op*2*PI/h), sin(-op*2*PI/h));
        for(int j = 0; j < n; 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;
            }
        }
    }
}

char a[N], b[N];
Complex A[N<<1], B[N<<1];
int ans[N<<1];
void poly_muilt(){
    int n = 1, len1 = strlen(a), len2 = strlen(b);
    while(n<len1*2 || n<len2*2)n<<=1;
    for(int i = 0; i < len1; i++)A[i] = a[len1-i-1]-'0';
    for(int i = len1; i < n; i++)A[i] = 0;
    for(int i = 0; i < len2; i++)B[i] = b[len2-i-1]-'0';
    for(int i = len2; i < n; i++)B[i] = 0;
    fft(A, n, 1);
    fft(B, n, 1);
    for(int i = 0; i < n; i++)
          A[i] *= B[i];
    fft(A, n, -1);
    for(int i = 0; i < n; i++)
          ans[i] = (int)(A[i].real()/n+0.5);
    for(int i = 0; i < n; i++){
        ans[i+1] += ans[i]/10;
        ans[i] %= 10;
    }
    n = len1+len2-1;
    while(ans[n] <= 0 && n > 0)n--;
    for(int i = n; i >= 0; i--)
      printf("%c", ans[i]+'0');
    printf("
");
}

int main()
{
    while(scanf("%s%s", a, b) != EOF){
        poly_muilt();
    }

    return 0;
}