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 = ;

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

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

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

     return ;
 }