[BZOJ-2179]FFT快速傅立叶

Description

给你2个高精度数,求这两个数的乘积。

Solution

60000的数据范围显然我们不能用O(n ^ 2)算法来解决。
所以我们把高精度数的每一位看成是一个系数,题目就转化成了多项式相乘。

Notice

注意最后的进位。

Code

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#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define sqz main
#define ll long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define Rep(i, a, b) for (int i = (a); i < (b); i++)
#define travel(i, u) for (int i = head[u]; ~i; i = edge[i].next)

const ll INF = 1e9, Mo = INF + 7;
const int N = 132000;
const double eps = 1e-6, pi = acos(-1.0);
ll read()
{
ll x = 0; int zf = 1; char ch = getchar();
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') zf = -1, ch = getchar();
while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
return x * zf;
}
void write(ll y)
{
if (y < 0) putchar('-'), y = -y;
if (y > 9) write(y / 10);
putchar(y % 10 + '0');
}

int T[N + 5], Rev[N + 5];
char s[N + 5];
struct Complex
{
double real, imagine;
Complex (double _real = 0, double _imagine = 0) { real = _real, imagine = _imagine;}
Complex operator +(const Complex X)
{
return Complex(real + X.real, imagine + X.imagine);
}
Complex operator -(const Complex X)
{
return Complex(real - X.real, imagine - X.imagine);
}
Complex operator *(const Complex X)
{
return Complex(real * X.real - imagine * X.imagine, real * X.imagine + imagine * X.real);
}
}X[N + 5], Y[N + 5];

void FFT(Complex *T, int n, int op)
{
Rep(i, 0, n) if (i > Rev[i]) swap(T[i], T[Rev[i]]);
for (int Size = 1; Size < n; Size <<= 1)
{
Complex wn = Complex(cos(pi / Size), sin(pi / Size) * op);
for (int L = 0; L < n; L += (Size << 1))
{
Complex w = Complex(1, 0);
Rep(R, L, L + Size)
{
Complex x = T[R], y = w * T[R + Size];
T[R] = x + y, T[R + Size] = x - y;
w = w * wn;
}
}
}
}

int sqz()
{
int n = read() - 1; int x = n * 2;
scanf("%s", s);
rep(i, 0, n) X[i] = s[n - i] - '0';
scanf("%s", s);
rep(i, 0, n) Y[i] = s[n - i] - '0';
int t, l = 0; for (t = 1; t <= x; t <<= 1) l++;
Rep(i, 0, t) Rev[i] = (Rev[i >> 1] >> 1) + ((i & 1) << l - 1);
FFT(X, t, 1); FFT(Y, t, 1);
Rep(i, 0, t) X[i] = X[i] * Y[i];
FFT(X, t, -1);
rep(i, 0, x) T[i] = (int)(X[i].real / t + 0.5);
Rep(i, 0, x) if (T[i] >= 10)
{
T[i + 1] += T[i] / 10;
T[i] %= 10;
}
if (T[x] >= 10) T[x + 1] += T[x] / 10, T[x] %= 10, x++;
per(i, x, 0) printf("%d", T[i]); puts("");
return 0;
}
文章目录
  1. 1. Description
  2. 2. Solution
  3. 3. Notice
  4. 4. Code