高精度乘高精度FFT优化算法

高精度乘高精度FFT优化算法

传入参数约定：传入参数均为string类型，返回值为string类型
算法思想：将两个高精度乘数每个数位上的数视为多项式对应的系数，用o(n*log(n))的复杂度转成点值形式，再利用o(n)的复杂度相乘，最后对点值进行差值，用o(n*log(n))的复杂度还原成多项式的形式，即原来的形式。
算法复杂度：o(n*log(n)).

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <map>#include <queue>#include <set>#include <vector>using namespace std;#define L(x) (1 << (x))const double PI = acos(-1.0);const int Maxn = 133015;double ax[Maxn], ay[Maxn], bx[Maxn], by[Maxn];char sa[Maxn/2],sb[Maxn/2];int sum[Maxn];int x1[Maxn],x2[Maxn];int revv(int x, int bits){    int ret = 0;    for (int i = 0; i < bits; i++)    {        ret <<= 1;        ret |= x & 1;        x >>= 1;    }    return ret;}void fft(double * a, double * b, int n, bool rev){    int bits = 0;    while (1 << bits < n) ++bits;    for (int i = 0; i < n; i++)    {        int j = revv(i, bits);        if (i < j)            swap(a[i], a[j]), swap(b[i], b[j]);    }    for (int len = 2; len <= n; len <<= 1)    {        int half = len >> 1;        double wmx = cos(2 * PI / len), wmy = sin(2 * PI / len);        if (rev) wmy = -wmy;        for (int i = 0; i < n; i += len)        {            double wx = 1, wy = 0;            for (int j = 0; j < half; j++)            {                double cx = a[i + j], cy = b[i + j];                double dx = a[i + j + half], dy = b[i + j + half];                double ex = dx * wx - dy * wy, ey = dx * wy + dy * wx;                a[i + j] = cx + ex, b[i + j] = cy + ey;                a[i + j + half] = cx - ex, b[i + j + half] = cy - ey;                double wnx = wx * wmx - wy * wmy, wny = wx * wmy + wy * wmx;                wx = wnx, wy = wny;            }        }    }    if (rev)    {        for (int i = 0; i < n; i++)            a[i] /= n, b[i] /= n;    }}int solve(int a[],int na,int b[],int nb,int ans[]){    int len = max(na, nb), ln;    for(ln=0; L(ln)<len; ++ln);    len=L(++ln);    for (int i = 0; i < len ; ++i)    {        if (i >= na) ax[i] = 0, ay[i] =0;        else ax[i] = a[i], ay[i] = 0;    }    fft(ax, ay, len, 0);    for (int i = 0; i < len; ++i)    {        if (i >= nb) bx[i] = 0, by[i] = 0;        else bx[i] = b[i], by[i] = 0;    }    fft(bx, by, len, 0);    for (int i = 0; i < len; ++i)    {        double cx = ax[i] * bx[i] - ay[i] * by[i];        double cy = ax[i] * by[i] + ay[i] * bx[i];        ax[i] = cx, ay[i] = cy;    }    fft(ax, ay, len, 1);    for (int i = 0; i < len; ++i)        ans[i] = (int)(ax[i] + 0.5);    return len;}string mul(string sa,string sb){    int l1,l2,l;    int i;    string ans;    memset(sum, 0, sizeof(sum));    l1 = sa.size();    l2 = sb.size();    for(i = 0; i < l1; i++)        x1[i] = sa[l1 - i - 1]-'0';    for(i = 0; i < l2; i++)        x2[i] = sb[l2-i-1]-'0';    l = solve(x1, l1, x2, l2, sum);    for(i = 0; i<l || sum[i] >= 10; i++) // 进位    {        sum[i + 1] += sum[i] / 10;        sum[i] %= 10;    }    l = i;    while(sum[l] <= 0 && l>0)    l--; // 检索最高位    for(i = l; i >= 0; i--)    ans+=sum[i] + '0'; // 倒序输出    return ans;}int main(){    cin.sync_with_stdio(false);    string a,b;    while(cin>>a>>b)  cout<<mul(a,b)<<endl;    return 0;}