Hdu 1402 （FFT）

题目链接

A * B Problem Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12490    Accepted Submission(s): 2206

Problem Description

Calculate A * B.

Input

Each line will contain two integers A and B. Process to end of file.

Note: the length of each integer will not exceed 50000.

Output

For each case, output A * B in one line.

Sample Input

1 2 1000 2

Sample Output

2 2000

初学FFT，根本不懂怎么实现的。直接套模板

Accepted Code:

#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <cctype>
#include <cassert>
#include <limits>
#include <bitset>
#include <complex>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define rer(i,l,u) for(int (i)=(int)(l);(i)<=(int)(u);++(i))
#define reu(i,l,u) for(int (i)=(int)(l);(i)<(int)(u);++(i))
#define all(o) (o).begin(), (o).end()
#define rall(o) (o).rbegin(), ((o).rend())
#define pb(x) push_back(x)
#define mp(x,y) make_pair((x),(y))
#define mset(m,v) memset(m,v,sizeof(m))
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3fLL
using namespace std;
typedef vector<int> vi; typedef pair<int,int> pii; typedef vector<pair<int,int> > vpii;
typedef long long ll; typedef vector<long long> vl; typedef pair<long long,long long> pll; typedef vector<pair<long long,long long> > vpll;
typedef vector<string> vs; typedef long double ld;
template<typename T, typename U> inline void amin(T &x, U y) { if(y < x) x = y; }
template<typename T, typename U> inline void amax(T &x, U y) { if(x < y) x = y; }

typedef long double Num;    //??????long double?????
const Num PI = 3.141592653589793238462643383279L;
typedef complex<Num> Complex;
//n?????
//a?????
void fft_main(int n, Num theta, Complex a[]) {
    for(int m = n; m >= 2; m >>= 1) {
        int mh = m >> 1;
        Complex thetaI = Complex(0, theta);
        rep(i, mh) {
            Complex w = exp((Num)i*thetaI);
            for(int j = i; j < n; j += m) {
                int k = j + mh;
                Complex x = a[j] - a[k];
                a[j] += a[k];
                a[k] = w * x;
            }
        }
        theta *= 2;
    }
    int i = 0;
    reu(j, 1, n-1) {
    for(int k = n >> 1; k > (i ^= k); k >>= 1) ;
        if(j < i) swap(a[i], a[j]);
    }
}

void fft(int n, Complex a[]) { fft_main(n, 2 * PI / n, a); }
void inverse_fft(int n, Complex a[]) { fft_main(n, -2 * PI / n, a); }

void convolution(vector<Complex> &v, vector<Complex> &w) {
    int n = 1, vwn = v.size() + w.size() - 1;
    while(n < vwn) n <<= 1;
    v.resize(n), w.resize(n);
    fft(n, &v[0]);
    fft(n, &w[0]);
    rep(i, n) v[i] *= w[i];
    inverse_fft(n, &v[0]);
    rep(i, n) v[i] /= n;
}

const int maxn = 50005;
vector<int> ans;
char s1[maxn], s2[maxn];

void solve(vector<int> &res) {    
    //cerr << s1 << s2;
    int len1 = strlen(s1), len2 = strlen(s2);
    for (int i = 0; i < len1; i++) s1[i] -= '0';
    for (int i = 0; i < len2; i++) s2[i] -= '0';
    vector<Complex> Lc(len1), Rc(len2);
    for (int i = 0; i < len1; i++) Lc[i] = Complex(s1[len1-i-1], 0);
    for (int i = 0; i < len2; i++) Rc[i] = Complex(s2[len2-i-1], 0);
    convolution(Lc, Rc);
    int n = len1 + len2 - 1;
    res.resize(n);
    rep(i, n) res[i] = Lc[i].real() + .5;
}

int main(void) {
    while (~scanf("%s %s", s1, s2)) {
        solve(ans);
        ans.resize(ans.size() << 1);
        for (int i = 0; i < ans.size(); i++) {
            ans[i+1] += ans[i] / 10;
            ans[i] %= 10;
        }
        int high = ans.size() - 1;
        while (high > 0 && !ans[high]) high--;
        for (int i = high; i >= 0; i--) putchar(ans[i] + '0');
        putchar('
');
    }
    return 0;
}