UVALive 6886 Golf Bot FFT

Golf Bot

题目连接：

http://acm.hust.edu.cn/vjudge/problem/visitOriginUrl.action?id=129724

Description

Do you like golf? I hate it. I hate golf so much that I

decided to build the ultimate golf robot, a robot that

will never miss a shot. I simply place it over the ball,

choose the right direction and distance and, flawlessly,

it will strike the ball across the air and into the hole.

Golf will never be played again.

Unfortunately, it doesn’t work as planned. So, here

I am, standing in the green and preparing my first

strike when I realize that the distance-selector knob

built-in doesn’t have all the distance options! Not everything

is lost, as I have 2 shots.

Given my current robot, how many holes will I be

able to complete in 2 strokes or less? The ball must be

always on the right line between the tee and the hole.

It isn’t allowed to overstep it and come back.

Input

The input file contains several test cases, each of them

as described below.

The first line has one integer: N, the number of

different distances the Golf Bot can shoot. Each of

the following N lines has one integer, ki

, the distance

marked in position i of the knob.

Next line has one integer: M, the number of holes in this course. Each of the following M lines has

one integer, dj , the distance from Golf Bot to hole j.

Constraints:

1 ≤ N, M ≤ 200 000

1 ≤ ki

, dj ≤ 200 000

Output

For each test case, you should output a single integer, the number of holes Golf Bot will be able to

complete. Golf Bot cannot shoot over a hole on purpose and then shoot backwards.

Sample Output Explanation

Golf Bot can shoot 3 different distances (1, 3 and 5) and there are 6 holes in this course at distances

2, 4, 5, 7, 8 and 9. Golf Bot will be able to put the ball in 4 of these:

• The 1st hole, at distance 2, can be reached by striking two times a distance of 1.

• The 2nd hole, at distance 4, can be reached by striking with strength 3 and then strength 1 (or

vice-versa).

• The 3rd hole can be reached with just one stroke of strength 5.

• The 5th hole can be reached with two strikes of strengths 3 and 5.

Holes 4 and 6 can never be reached

Sample Input

3

1

3

5

6

2

4

5

7

8

9

Sample Output

4

Hint

题意

给你n个数，然后再给你一个数k,问这个数是否就是那n个数中的一个，或者说这个数可以由这n个数中的两个构成（可以是自己*2）

题解：

裸的不行的FFT，直接做就好了。

代码

#include<bits/stdc++.h>

using namespace std;

const int N = 1200040;
const double pi = acos(-1.0);

int len;

struct Complex
{
    double r,i;
    Complex(double r=0,double i=0):r(r),i(i) {};
    Complex operator+(const Complex &rhs)
    {
        return Complex(r + rhs.r,i + rhs.i);
    }
    Complex operator-(const Complex &rhs)
    {
        return Complex(r - rhs.r,i - rhs.i);
    }
    Complex operator*(const Complex &rhs)
    {
        return Complex(r*rhs.r - i*rhs.i,i*rhs.r + r*rhs.i);
    }
} va[N],vb[N];

void rader(Complex F[],int len) //len = 2^M,reverse F[i] with  F[j] j为i二进制反转
{
    int j = len >> 1;
    for(int i = 1;i < len - 1;++i)
    {
        if(i < j) swap(F[i],F[j]);  // reverse
        int k = len>>1;
        while(j>=k)
        {
            j -= k;
            k >>= 1;
        }
        if(j < k) j += k;
    }
}

void FFT(Complex F[],int len,int t)
{
    rader(F,len);
    for(int h=2;h<=len;h<<=1)
    {
        Complex wn(cos(-t*2*pi/h),sin(-t*2*pi/h));
        for(int j=0;j<len;j+=h)
        {
            Complex E(1,0); //旋转因子
            for(int k=j;k<j+h/2;++k)
            {
                Complex u = F[k];
                Complex v = E*F[k+h/2];
                F[k] = u+v;
                F[k+h/2] = u-v;
                E=E*wn;
            }
        }
    }
    if(t==-1)   //IDFT
        for(int i=0;i<len;++i)
            F[i].r/=len;
}

void Conv(Complex a[],Complex b[],int len) //求卷积
{
    FFT(a,len,1);
    FFT(b,len,1);
    for(int i=0;i<len;++i) a[i] = a[i]*b[i];
    FFT(a,len,-1);
}
int n;
int a[N];
long long num[N],sum[N];
void solve()
{
    memset(num,0,sizeof(num));
    memset(sum,0,sizeof(sum));
    memset(va,0,sizeof(va));
    memset(vb,0,sizeof(vb));
    int Mx = 0;
    for(int i=0;i<n;i++)
    {
        int x;scanf("%d",&a[i]);
        Mx = max(Mx,a[i]);
        num[a[i]]=1;
    }
    Mx*=2;
    len=1;
    while(len<=Mx+1)len*=2;
    sort(a,a+n);
    for(int i=0;i<=len;i++)
    {
        va[i].r=num[i];
        va[i].i=0;
        vb[i].r=va[i].r;
        vb[i].i=0;
    }
    Conv(va,vb,len);
    for(int i=0;i<len;i++)
        num[i]+=(long long)(va[i].r+0.5);
    int cnt = 0;
    int q;scanf("%d",&q);
    while(q--){
        int bbb;
        scanf("%d",&bbb);
        if(num[bbb])cnt++;
    }
    printf("%d\n",cnt);
}
int main()
{
    while(scanf("%d",&n)!=EOF)solve();
    return 0;
}