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
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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

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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