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 = 5000005;
const double pi = acos(-1);
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 = 0; i < n; ++i)
        if (i < rev[i])swap(c[i], c[rev[i]]);

    for (int i = 1; i < n; i <<= 1)
    {
        Complex wn(cos(pi/i), f*sin(pi/i));

        for (int j = 0; j < n; j += (i << 1))
        {
            Complex wk(1, 0);

            for (int k = 0; 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 = 0; i < n; ++i)
        a[i] = str[n - i - 1] - '0';

    scanf("%s", str);

    for (int i = 0; i < n; ++i)
        b[i] = str[n - i - 1] - '0';

    m = n << 1;
    for (n = 1; n < m; )
        ++len, n <<= 1;
    
    for (int i = 0; i < n; ++i)
    {
        rev[i] |= rev[i >> 1] >> 1;
        rev[i] |= (i&1) << (len - 1);
    }

    calculateFFT(a, 1);
    calculateFFT(b, 1);

    for (int i = 0; i < n; ++i)
        a[i] *= b[i];

    calculateFFT(a, -1);

    for (int i = 0; i < n; ++i)
        a[i] /= n;

    for (int i= 0; i < m; ++i)
        ans[i] = int(a[i].real() + 0.5);

    while (!ans[m - 1])--m;

    for (int i = 0; i < m; ++i)
        if (ans[i] >= 10)
        {
            ans[i + 1] += ans[i]/10;
            ans[i] %= 10;
        }

    if (!ans[m])--m;

    for (int i = m; ~i; --i)
        putchar('0' + ans[i]);
}

@Author: YouSiki