UVALive - 6886 Golf Bot 多项式乘法（FFT）

题目链接：

http://acm.hust.edu.cn/vjudge/problem/129724

Golf Bot

Time Limit: 15000MS

题意

给你n个数，m个查询，对于每个查询，问能不能用n个数中的一个或两个（同一个数可以取两次）相加凑出来。

题解

多项式乘法,用快速傅里叶变化加速，时间复杂度：O(nlogn)。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

const int maxn=555555;
int vis[maxn],n;

struct Complex {
	double real, image;
	Complex(double real, double image):real(real),image(image) {}
	Complex() {}
	friend Complex operator + (const Complex &c1, const Complex &c2) {
		return Complex(c1.real + c2.real, c1.image + c2.image);
	}
	friend Complex operator - (const Complex &c1, const Complex &c2) {
		return Complex(c1.real - c2.real, c1.image - c2.image);
	}
	friend 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);
	}
}a[maxn];

struct IterativeFFT {
	Complex A[maxn];

	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;
	}

	//当DFT= 1时是DFT, DFT = -1则是逆DFT
	//对长度为len(2的幂)的数组进行DFT变换
	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));
			//这一层结点的包含数组元素个数都是(1 << s)
			for(int k = 0; k < len; k += m) {
				Complex w = Complex(1, 0);
				//折半引理, 根据两个子节点计算父亲节点
				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];
	}
} myfft;

void init() {
	rep(i,0,maxn) a[i]=Complex(0,0);
	a[0].real=1;
	clr(vis,0);
}

int main() {
	while(scanf("%d",&n)==1&&n) {
		init();

		int ma=0;
		rep(i,0,n){
			int x;
			scanf("%d",&x);
			ma=max(ma,x);
			a[x].real=1.0;
		}

		int len=1;
		while(len<ma) len<<=1;
		len<<=1;
		myfft.FFT(a, len, 1);
		rep(i,0,len){
			a[i]=a[i]*a[i];
		}
		myfft.FFT(a,len,-1);
		rep(i,0,len){
			if(a[i].real>eps) vis[i]=1;
		}

		int q;
		scanf("%d",&q);
		int ans=0;
		while(q--) {
			int x;
			scanf("%d",&x);
			if(vis[x]) {
				ans++;
			}
		}

		printf("%d\n",ans);
	}
	return 0;
}

这题用bitset也能做，不过跑的没有FFT快。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=222223;
bitset<maxn> bset1,bset2;
int arr[maxn];
int n;

int main() {
    while(scanf("%d",&n)==1&&n){
    	bset1.reset();
    	rep(i,0,n){
    		scanf("%d",&arr[i]);
			bset1.set(arr[i]); 
		}
		bset2=bset1;
		rep(i,0,n){
			bset1|=bset2<<arr[i];
		}
		int q; scanf("%d",&q);
		int ans=0;
		while(q--){
			int x; scanf("%d",&x);
			if(bset1.test(x)) ans++;
		}
		printf("%d\n",ans);
	}
    return 0;
}

//end-----------------------------------------------------------------------