UVALive 6886 (LA 6886) Golf Bot FFT

题目大意：

就是现在给出N个数, 然后给出M个数询问这M书中有多少个可以是由这N个数中两个相加(同一个数可选两次)组成或者等于这N个数中的某一个

大致思路：

就是构造多项式相乘就可以了, FFT的最简单的应用了, 将N个数a1, a2,...an对应成多项式, 用x^k的系数是1表示N个数中有值为k的, 否则系数用0表示

然后求这个多项式的平方就可以了

代码如下：

Result  :  Accepted     Memory  :  0 KB     Time  :  785 ms

/*
 * Author: Gatevin
 * Created Time:  2015/7/17 14:08:34
 * File Name: UVALive6886.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

const double PI = acos(-1.0);

struct Complex
{
    double real, image;
    Complex(double _real, double _image)
    {
        real = _real;
        image = _image;
    }
    Complex(){}
};

Complex operator + (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real + c2.real, c1.image + c2.image);
}

Complex operator - (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real - c2.real, c1.image - c2.image);
}

Complex operator * (const Complex &c1, const Complex &c2)
{
    return Complex(c1.real*c2.real - c1.image*c2.image, c1.real*c2.image + c1.image*c2.real);
}

int rev(int id, int len)
{
    int ret = 0;
    for(int i = 0; (1 << i) < len; i++)
    {
        ret <<= 1;
        if(id & (1 << i)) ret |= 1;
    }
    return ret;
}

Complex A[1 << 19];
void FFT(Complex *a, int len, int DFT)
{
    for(int i = 0; i < len; i++)
        A[rev(i, len)] = a[i];
    for(int s = 1; (1 << s) <= len; s++)
    {
        int m = (1 << s);
        Complex wm = Complex(cos(DFT*2*PI/m), sin(DFT*2*PI/m));
        for(int k = 0; k < len; k += m)
        {
            Complex w = Complex(1, 0);
            for(int j = 0; j < (m >> 1); j++)
            {
                Complex t = w*A[k + j + (m >> 1)];
                Complex u = A[k + j];
                A[k + j] = u + t;
                A[k + j + (m >> 1)] = u - t;
                w = w*wm;
            }
        }
    }
    if(DFT == -1) for(int i = 0; i < len; i++) A[i].real /= len, A[i].image /= len;
    for(int i = 0; i < len; i++) a[i] = A[i];
    return;
}

Complex dis[1 << 19];
bool ok[200010];

int main()
{
    int n, m;
    while(~scanf("%d", &n))
    {
        memset(ok, 0, sizeof(ok));
        int maxDis = 0;
        int tmp;
        for(int i = 0; i < n; i++)
            scanf("%d", &tmp), ok[tmp] = 1, maxDis = max(maxDis, tmp);
        for(int i = 0; i <= maxDis; i++)
            if(ok[i]) dis[i] = Complex(1, 0);
            else dis[i] = Complex(0, 0);
        int len = 1;
        while(len <= maxDis) len <<= 1;
        len <<= 1;
        for(int i = maxDis + 1; i < len; i++)
            dis[i] = Complex(0, 0);
        FFT(dis, len, 1);
        for(int i = 0; i < len; i++)
            dis[i] = dis[i]*dis[i];
        FFT(dis, len, -1);
        int ans = 0;
        scanf("%d", &m);
        for(int i = 0; i < m; i++)
        {
            scanf("%d", &tmp);
            if(tmp <= maxDis*2 && (dis[tmp].real > 0.5 || ok[tmp]))
                ans++;
        }
        printf("%d\n", ans);
        
    }
    return 0;
}