2016级算法第六次上机-B.ModricWang's FFT : EASY VERSION

1114 ModricWang's FFT EASY VERSION

思路

利用FFT做大整数乘法，实际上是把大整数变成多项式，然后做多项式乘法。

例如，对于\(1234\)，改写成\(f(x)=1*x^3+2*x^2+3*x+4\)，那么\(x=10\)处的值就是原数。类似的，对于输入的两个大整数，转换为\(f(x)\) 和\(g(x)\) ，利用FFT求出\(h(x)=f(x)*g(x)\) ，此时\(h(10)\) 就是乘积。

代码

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <random>
#include <functional>
#include <complex>

using namespace std;
const auto PI = acos(-1.0);
typedef complex<double> Complex;

void change(Complex y[], int len) {
    int i, j, k;
    for (i = 1, j = len / 2; i < len - 1; i++) {
        if (i < j) swap(y[i], y[j]);
        k = len / 2;
        while (j >= k) {
            j -= k;
            k /= 2;
        }
        if (j < k)j += k;
    }
}

void fft(Complex y[], int len, int on) {
    change(y, len);
    for (int h = 2; h <= len; h <<= 1) {
        Complex wn(cos(-on * 2 * PI / h), sin(-on * 2 * PI / h));
        for (int j = 0; j < len; 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;
            }
        }
    }
    if (on == -1)
        for (int i = 0; i < len; i++)
            y[i] = Complex(y[i].real() / len, y[i].imag());
}

const int MAXN = 200010;
Complex x1[MAXN], x2[MAXN];
string str1, str2;
int sum[MAXN];

int main() {
#ifdef ONLINE_JUDGE
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
#endif
    cin >> str1 >> str2;
    auto len1 = str1.length();
    auto len2 = str2.length();
    auto len = 1;
    while (len < len1 * 2 || len < len2 * 2)len <<= 1;
    for (auto i = 0; i < len1; i++)
        x1[i] = Complex(str1[len1 - 1 - i] - '0', 0);
    for (auto i = len1; i < len; i++)
        x1[i] = Complex(0, 0);
    for (int i = 0; i < len2; i++)
        x2[i] = Complex(str2[len2 - 1 - i] - '0', 0);
    for (auto i = len2; i < len; i++)
        x2[i] = Complex(0, 0);

    fft(x1, len, 1);
    fft(x2, len, 1);
    for (auto i = 0; i < len; i++)
        x1[i] = x1[i] * x2[i];
    fft(x1, len, -1);
    for (int i = 0; i < len; i++)
        sum[i] = static_cast<int>(lround(x1[i].real()));
    for (int i = 0; i < len; i++) {
        sum[i + 1] += sum[i] / 10;
        sum[i] %= 10;
    }
    len = len1 + len2 - 1;
    while (sum[len] <= 0 && len > 0)len--;
    for (int i = len; i >= 0; i--)
        cout << static_cast<char>(sum[i] + '0');
    cout << "\n";

    return 0;
}