BZOJ 2179: FFT快速傅立叶

2179: FFT快速傅立叶

Time Limit: 10 Sec  Memory Limit: 259 MB
Submit: 2923  Solved: 1498
[Submit][Status][Discuss]

Description

给出两个n位10进制整数x和y，你需要计算x*y。

Input

第一行一个正整数n。
第二行描述一个位数为n的正整数x。
第三行描述一个位数为n的正整数y。

Output

输出一行，即x*y的结果。

Sample Input

1
3
4

Sample Output

12

数据范围：
n<=60000

HINT

Source

[Submit][Status][Discuss]

FFT模板

 #include <bits/stdc++.h>
 
 using namespace std;
 
 const int maxn = ;
 const double pi = acos(-);
 typedef complex<double> Complex;
 
 int n, m, len;
 int ans[maxn];
 int rev[maxn];
 char str[maxn];
 
 Complex a[maxn];
 Complex b[maxn];
 
 inline void calculateFFT(Complex *c, double f)
 {
     for (int i = ; i < n; ++i)
         if (i < rev[i])swap(c[i], c[rev[i]]);
 
     for (int i = ; i < n; i <<= )
     {
         Complex wn(cos(pi/i), f*sin(pi/i));
 
         for (int j = ; j < n; j += (i << ))
         {
             Complex wk(, );
 
             for (int k = ; k < i; ++k, wk *= wn)
             {
                 Complex x = c[j + k];
                 Complex y = c[i + j + k] * wk;
                 c[j + k] = x + y;
                 c[i + j + k] = x - y;
             }
         }
     }
 }
 
 signed main(void)
 {
     scanf("%d", &n);
 
     scanf("%s", str);
 
     for (int i = ; i < n; ++i)
         a[i] = str[n - i - ] - '';
 
     scanf("%s", str);
 
     for (int i = ; i < n; ++i)
         b[i] = str[n - i - ] - '';
 
     m = n << ;
     for (n = ; n < m; )
         ++len, n <<= ;
 
     for (int i = ; i < n; ++i)
     {
         rev[i] |= rev[i >> ] >> ;
         rev[i] |= (i&) << (len - );
     }
 
     calculateFFT(a, );
     calculateFFT(b, );
 
     for (int i = ; i < n; ++i)
         a[i] *= b[i];
 
     calculateFFT(a, -);
 
     for (int i = ; i < n; ++i)
         a[i] /= n;
 
     for (int i= ; i < m; ++i)
         ans[i] = int(a[i].real() + 0.5);
 
     while (!ans[m - ])--m;
 
     for (int i = ; i < m; ++i)
         if (ans[i] >= )
         {
             ans[i + ] += ans[i]/;
             ans[i] %= ;
         }
 
     if (!ans[m])--m;
 
     for (int i = m; ~i; --i)
         putchar('' + ans[i]);
 }

@Author: YouSiki