# CC countari & 分块+FFT

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
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(){
block=3;
if ((int)n/sqrt(n)>3) block=(int)n/sqrt(n);
FORP(i,1,n) {
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
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(){
block=4000;
//if ((ll)n/sqrt(n)>3) block=(ll)n/sqrt(n);
block=min(block,n);
FORP(i,1,n) {
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