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;
}