CC countari & 分块+FFT

题意:

  求一个序列中顺序的长度为3的等差数列.

SOL:

  对于这种计数问题都是用个数的卷积来进行统计.然而对于这个题有顺序的限制,不好直接统计,于是竟然可以分块?惊为天人...

  考虑分块以后的序列:

    一个块内直接枚举统计三个或两个在块内的.

    只有一个在当前块我们假设它是中间那个,对左右其它块做卷积.

  但是还是感觉复杂度有点玄学啊...

  我比较傻逼...一开始块内统计根本没有想清楚...最后做卷积硬生生把复杂度变成了 $\sqrt{N}*N*log(N)$...

  改了一个晚上终于没忍住看标程...我是傻逼明明a的范围比n小...

CODE:(tle的代码不想改了= =)

  

/*=================================
# Created time: 2016-04-19 16:09
# Filename: cccountari.cpp
# Description: Ploblem from codechef countari
=================================*/
#define me AcrossTheSky&HalfSummer11  
#include <cstdio>  
#include <cmath>  
#include <ctime>  
#include <string>  
#include <cstring>  
#include <cstdlib>  
#include <iostream>  
#include <algorithm>  
  
#include <set> 
#include <stack>  
#include <queue>  
#include <vector>  
  
#define lowbit(x) (x)&(-x)  
#define Abs(x) ((x) > 0 ? (x) : (-(x)))  
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)  
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)  
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)  
#define ls(a,b) (((a)+(b)) << 1)  
#define rs(a,b) (((a)+(b)) >> 1)  
#define getlc(a) ch[(a)][0]  
#define getrc(a) ch[(a)][1]  
  
#define maxn 500005 
#define maxm 100005 
#define INF 1070000000  
using namespace std;  
typedef long long ll;  
typedef unsigned long long ull;  
  
template<class T> inline  
void read(T& num){  
    num = 0; bool f = true;char ch = getchar();  
    while(ch < '0' || ch > '9') { if(ch == '-') f = false;ch = getchar();}  
    while(ch >= '0' && ch <= '9') {num = num * 10 + ch - '0';ch = getchar();}  
    num = f ? num: -num;  
} 
int outs[100]; 
template<class T> inline 
void write(T x){ 
	if (x==0) {putchar('0'); putchar(' '); return;} 
	if (x<0) {putchar('-'); x=-x;} 
	int num=0; 
	while (x){ outs[num++]=(x%10); x=x/10;} 
	FORM(i,num-1,0) putchar(outs[i]+'0'); putchar(' '); 
} 
/*==================split line==================*/
const double pi=acos(-1);
struct cpx{
    double x,y;
    inline double real(){return x;}
    cpx(double a=0,double b=0):x(a),y(b){}
}f[maxn],g[maxn],eps[maxn],inv_eps[maxn];

inline cpx operator +(cpx a,cpx b){return cpx(a.x+b.x,a.y+b.y);}
inline cpx operator -(cpx a,cpx b){return cpx(a.x-b.x,a.y-b.y);}
inline cpx operator /(cpx a,double b){return cpx(a.x/b,a.y);}
inline cpx operator *(cpx a,cpx b){return cpx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
inline cpx conj(cpx a){return cpx(a.x,-a.y);}

inline void get_eps(int p){
	double angle=2.0*pi/p;
	FORP(i,0,p-1) eps[i]=cpx(cos(angle*i),sin(angle*i));
	FORP(i,0,p-1) inv_eps[i]=conj(eps[i]);
}
inline void fft(int n,cpx* buffer,cpx* eps){
	for (int i = 0, j = 0;i < n; i++){
		if (i > j) swap(buffer[i],buffer[j]);
		for (int l = n >> 1;(j ^= l) < l;l >>= 1);
	}
	for (int i = 2; i <= n; i <<= 1){
		int m = i >> 1;
		for (int j = 0; j < n; j += i)
			for (int k = 0; k != m; k++){
				cpx z = buffer[j + m + k] * eps[n / i * k];
				buffer[j + m + k] = buffer[j + k] - z;
				buffer[j + k] = buffer[j + k] + z;
			}
	}
}
ll n,k,block,num;
ll prec[30004],c[30040],suf[40000];
//bool vis[40000];
ll ans=0;
ll maxa=0;
struct Range{ ll l,r; } rg[1000];
struct Infor{ ll val,pos;} a[100005];

int main(){
	read(n);
	block=3;
	if ((int)n/sqrt(n)>3) block=(int)n/sqrt(n);
	FORP(i,1,n) {
		read(a[i].val);
		g[a[i].val].x++;
		maxa=max(maxa,a[i].val);
		a[i].pos=(i-1)/block+1;
		suf[a[i].val]++;
	}
	
	ll k=1; maxa*=2;
	while (k<maxa) k<<=1;
	k<<=1;
	get_eps(k);
	
	num=a[n].pos;
	FORP(i,1,num) rg[i].l=(i-1)*block+1,rg[i].r=min(n,(i)*block);
	FORP(i,1,num){
		FORP(j,rg[i].l,rg[i].r) c[a[j].val]++;
		FORP(l,rg[i].l,rg[i].r-1){
			suf[a[l].val]--;
			FORP(r,l+1,rg[i].r) {
				suf[a[r].val]--;
				if (a[l].val==a[r].val) ans+=suf[a[l].val]+prec[a[l].val];
				else
				{
					int x=a[l].val-(a[r].val-a[l].val);
					if (x>0) ans+=prec[x];
					x=a[r].val+(a[r].val-a[l].val);
					if (x>0) ans+=suf[x];
				}
			}
			FORP(r,l+1,rg[i].r)	suf[a[r].val]++;
		}
		FORP(j,rg[i].l,rg[i].r) prec[a[j].val]+=c[a[j].val],c[a[j].val]=0;
		suf[a[rg[i].r].val]--;
	}
	FORP(i,rg[1].l,rg[1].r) g[a[i].val].x--;
	FORP(i,2,num-1){
		//memset(f,0,sizeof(f)); memset(g,0,sizeof(g));
		FORP(j,rg[i-1].l,rg[i-1].r) f[a[j].val].x++;
		FORP(j,rg[i-1].l,rg[i-1].r) g[a[j].val].x--;
		fft(k,f,eps); fft(k,g,eps);
		FORP(j,0,k-1) f[j]=f[j]*g[j];
		fft(k,f,inv_eps);
		FORP(j,rg[i].l,rg[i].r) ans+=(ll)(f[a[j].val+a[j].val].x/k+0.5);
	}
	write(ans); puts("");
}

 调下块大小a掉了

/*=================================
# Created time: 2016-04-19 16:09
# Filename: cccountari.cpp
# Description: Ploblem from codechef countari
=================================*/
#define me AcrossTheSky&HalfSummer11  
#include <cstdio>  
#include <cmath>  
#include <ctime>  
#include <string>  
#include <cstring>  
#include <cstdlib>  
#include <iostream>  
#include <algorithm>  
  
#include <set> 
#include <stack>  
#include <queue>  
#include <vector>  
  
#define lowbit(x) (x)&(-x)  
#define Abs(x) ((x) > 0 ? (x) : (-(x)))  
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)  
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)  
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)  
#define ls(a,b) (((a)+(b)) << 1)  
#define rs(a,b) (((a)+(b)) >> 1)  
#define getlc(a) ch[(a)][0]  
#define getrc(a) ch[(a)][1]  
  
#define maxn 100005 
#define maxm 100005 
#define INF 1070000000  
using namespace std;  
typedef long long ll;  
typedef unsigned long long ull;  
  
template<class T> inline  
void read(T& num){  
    num = 0; bool f = true;char ch = getchar();  
    while(ch < '0' || ch > '9') { if(ch == '-') f = false;ch = getchar();}  
    while(ch >= '0' && ch <= '9') {num = num * 10 + ch - '0';ch = getchar();}  
    num = f ? num: -num;  
} 
int outs[100]; 
template<class T> inline 
void write(T x){ 
	if (x==0) {putchar('0'); putchar(' '); return;} 
	if (x<0) {putchar('-'); x=-x;} 
	int num=0; 
	while (x){ outs[num++]=(x%10); x=x/10;} 
	FORM(i,num-1,0) putchar(outs[i]+'0'); //putchar(' '); 
} 
/*==================split line==================*/
const double pi=acos(-1);
struct cpx{
    double x,y;
    cpx(double a=0,double b=0):x(a),y(b){}
}f[maxn],g[maxn],eps[maxn],inv_eps[maxn];

inline cpx operator +(cpx a,cpx b){return cpx(a.x+b.x,a.y+b.y);}
inline cpx operator -(cpx a,cpx b){return cpx(a.x-b.x,a.y-b.y);}
inline cpx operator *(cpx a,cpx b){return cpx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
inline cpx conj(cpx a){return cpx(a.x,-a.y);}

inline void get_eps(int p){
	double angle=2.0*pi/p;
	FORP(i,0,p-1) eps[i]=cpx(cos(angle*i),sin(angle*i));
	FORP(i,0,p-1) inv_eps[i]=conj(eps[i]);
}
inline void fft(int n,cpx* buffer,cpx* eps){
	for (int i = 0, j = 0;i < n; i++){
		if (i > j) swap(buffer[i],buffer[j]);
		for (int l = n >> 1;(j ^= l) < l;l >>= 1);
	}
	for (int i = 2; i <= n; i <<= 1){
		int m = i >> 1;
		for (int j = 0; j < n; j += i)
			for (int k = 0; k != m; k++){
				cpx z = buffer[j + m + k] * eps[n / i * k];
				buffer[j + m + k] = buffer[j + k] - z;
				buffer[j + k] = buffer[j + k] + z;
			}
	}
}
int n,k,block,num;
int prec[30010],c[30010],suf[30100];
ll ans=0;
int maxa=0;
struct Range{ int l,r; } rg[10000];
struct Infor{ int val,pos;} a[maxn];

int main(){
	read(n);
	block=4000;
	//if ((ll)n/sqrt(n)>3) block=(ll)n/sqrt(n);
	block=min(block,n);
	FORP(i,1,n) {
		read(a[i].val);
		maxa=max(maxa,a[i].val);
		a[i].pos=(i-1)/block+1;
		suf[a[i].val]++;
	}
	
	int k=1; //maxa*=2;
	while (k<maxa) k<<=1;
	k<<=1;
	get_eps(k);
	
	num=a[n].pos;
	FORP(i,1,num) rg[i].l=(i-1)*block+1,rg[i].r=(i)*block; rg[num].r=n;
	FORP(i,1,num){
		FORP(j,rg[i].l,rg[i].r) c[a[j].val]++;
		FORP(l,rg[i].l,rg[i].r-1){
			suf[a[l].val]--;
			FORP(r,l+1,rg[i].r) {
				suf[a[r].val]--;
				if (a[l].val==a[r].val) ans+=suf[a[l].val]+prec[a[l].val];
				else
				{
					int x=a[l].val-(a[r].val-a[l].val);
					if (x>0 && x<=maxa) ans+=prec[x];
					x=a[r].val+(a[r].val-a[l].val);
					if (x>0 && x<=maxa) ans+=suf[x];
				}
			}
			FORP(r,l+1,rg[i].r)	suf[a[r].val]++;
		}
		suf[a[rg[i].r].val]--;
		FORP(j,0,maxa) f[j]=cpx(prec[j],0),g[j]=cpx(suf[j],0);
		FORP(j,maxa+1,k) f[j]=cpx(0.0,0),g[j]=cpx(0.0,0);
		fft(k,f,eps); fft(k,g,eps);
		FORP(j,0,k-1) f[j]=f[j]*g[j];
		fft(k,f,inv_eps);
		FORP(j,rg[i].l,rg[i].r) ans+=trunc(f[a[j].val+a[j].val].x/k+0.5);
		
		FORP(j,rg[i].l,rg[i].r) prec[a[j].val]+=c[a[j].val],c[a[j].val]=0;
		
	}
	/*FORP(i,2,num-1){
		memset(f,0,sizeof(f)); memset(g,0,sizeof(g));
		FORP(j,1,rg[i].l-1) f[a[j].val].x++;
		FORP(j,rg[i].r+1,n) g[a[j].val].x++;
		fft(k,f,eps); fft(k,g,eps);
		FORP(j,0,k-1) f[j]=f[j]*g[j];
		fft(k,f,inv_eps);
		//FORP(j,0,maxa) printf("%d ",(int)(f[j].x/k+0.5)); puts("");
		FORP(j,rg[i].l,rg[i].r) ans+=(int)(f[a[j].val+a[j].val].x/k+0.5);
	}*/
	write(ans); puts("");
	//printf("%lf",clock()-t);
}

 

posted @ 2016-04-20 07:37  YCuangWhen  阅读(391)  评论(0编辑  收藏