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hdu1402 A * B Problem Plus

很裸的FFT。

原理到處都有(看演算法導論就可以學懂了)。

#include<bits/stdc++.h>
using namespace std;

typedef complex<double> cpx;

const int MAXN = 50000 * 3;
const double PI = 3.1415926535897932384626433832795;

void FFT(vector<cpx> &A, bool inv)  {
    int n = A.size();
    for(int i = 0, j = 0; i < n; i++)    {
        if(j > i)   swap(A[j], A[i]);
        int t = n;
        while(j & (t >>= 1))
            j ^= t;
        j |= t;
    }
    double pi = inv ? -PI : PI;
    for(int step = 1; step < n; step <<= 1) {
        double alpha = pi / step;
        for(int k = 0; k < step; k++)  {
            cpx wk = exp(cpx(0, alpha * k));
            for(int Ek = k; Ek < n; Ek += step << 1)    {
                int Ok = Ek + step;
                cpx t = wk * A[Ok];
                A[Ok] = A[Ek] - t;
                A[Ek] = A[Ek] + t;
            }
        }
    }
    if(inv) for(int i = 0; i < n; i++)
        A[i] /= n;
}

void Multi(vector<int> &a, vector<int> &b, vector<int> &c)    {
    int len = 1, M = max(a.size(), b.size()) * 2;
    while(len < M)  len <<= 1;

    vector<cpx> A(len, 0), B(len, 0);
    for(int i = 0; i < a.size(); i++)   A[i].real(a[i]);
    for(int i = 0; i < b.size(); i++)   B[i].real(b[i]);

    FFT(A, 0), FFT(B, 0);
    for(int i = 0; i < len; i++)    A[i] *= B[i];
    FFT(A, 1);

    c.resize(len);
    long long over = 0;
    for(int i = 0; i < len; i++)    {
        over += A[i].real() + 0.5;
        c[i] = over % 10;
        over /= 10;
    }
    while(!c[len - 1] && len > 1)
        len--;
    c.resize(len);
}

vector<int> a, b, c;
string s;

int main()  {
    while(cin >> s) {
        a.clear(), b.clear();
        for(int i = 1; i <= s.size(); i++)  a.push_back(s[s.size() - i] - '0');
        cin >> s;
        for(int i = 1; i <= s.size(); i++)  b.push_back(s[s.size() - i] - '0');
        Multi(a, b, c);
        for(int i = c.size() - 1; i >= 0; i--)
            printf("%d", c[i]);
        puts("");
    }
    return 0;
}