UVALive - 6886 Golf Bot 多项式乘法（FFT）
题目链接：
http://acm.hust.edu.cn/vjudge/problem/129724
Golf Bot
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-----------------------------------------------------------------------
