【BZOJ4927】第一题 双指针+DP

题解：

虽然是过了，不过做的十分智障

首先是有 2根 2 1 1 ， 3根 1 1 1

这两种方法

然后考虑2 2 1 1

two-point-two没啥好说的

3 1 1 1

我很智障的以为数据范围是1e9

然后写了hash，刚开始还开错范围

把int换short int才卡过了空间

首先枚举 1 1 1

然后枚举3 中的一条边

既然1e7直接记录每个数能由两个数组成的方案数

另外再减去这条边参与的方案数（直接两个数减一下查有几个就好了）

然后细节还挺多的搞个对拍就可以了

#pragma G++ optimize (2)
#include <bits/stdc++.h>
using namespace std;
#define N 6000
#define rg register
#define IL inline
#define ll long long
int a[N],n;
unsigned ll ans,ans3;
#define js(x) x*(x-1)/2*(x-2)/3*(x-3)/4
const int mo1=1.3e7+;
const int NN=1.3e7+1e5;
const int mo2=2.6e7+;
const int N2=2.6e7+1e5;
int hash1[NN],hash2[NN],hash4[N2];
short int hash3[N2],hash5[N2];
char ss[<<],*A=ss,*B=ss;
inline char gc(){return A==B&&(B=(A=ss)+fread(ss,,<<,stdin),A==B)?EOF:*A++;}
template<class T>void read(T&x){
    rg int f=,c;while(c=gc(),c<||<c)if(c=='-')f=-;x=c^;
    while(c=gc(),<c&&c<)x=(x<<)+(x<<)+(c^);x*=f;
}
IL void push(int x)
{
  int y=x%mo1;
  while (hash1[y]&&hash1[y]!=x) y++;
  hash1[y]=x; hash2[y]++;
}
IL int find1(int x)
{
  int y=x%mo1;
  while (hash1[y]&&hash1[y]!=x) y++;
  if (hash1[y]==x) return(hash2[y]);
  else return();
}
IL void push2(int x,int y)
{
  int z=(1ll*y*+x)%mo2;
  while (hash3[z]&&!(hash3[z]==x&&hash4[z]==y)) z++;
  hash3[z]=x; hash4[z]=y; hash5[z]++;
}
IL int find2(int x,int y)
{
  int z=(1ll*y*+x)%mo2;
  while (hash3[z]&&!(hash3[z]==x&&hash4[z]==y)) z++;
  if (hash3[z]==x&&hash4[z]==y) return(hash5[z]);
  else return();
}
IL bool cmp(int a,int b)
{
  return(a<b);
}
int now,cnt;
#define INF 1e9
int main()
{
  freopen("noi.in","r",stdin);
  freopen("noi.out","w",stdout);
  read(n);
  for (rg int i=;i<=n;i++) read(a[i]);
  sort(a+,a+n+,cmp);
  for (rg int i=;i<=n;i++)
    for (rg int j=i+;j<=n;j++)
      push(a[i]+a[j]);
  a[]=INF;
  for (rg int i=;i<=n;i++)
      for (rg int j=i+;j<=n;j++)
          push2(i,a[i]+a[j]),push2(j,a[i]+a[j]);
  now=;
  while (++now<=n)
  {
    cnt=;
    while (now<n&&a[now+]==a[now]) cnt++,now++;
    if (cnt>=)
    {
      int h=,t=n;
      ll ans2=;
      while (h<t)
      {
        int h1=h,t1=t;
        while (a[h]==a[h+]&&h<t) h++;
        while (a[t]+a[h]>a[now]&&h<t) t--;
        if (a[t]+a[h]!=a[now])
        {
          h++;
          continue;
        }
        t1=t;
        while (a[t]==a[t-]&&h<t) t--;
        if (h==t)
        {
          int xx=t1-h1+;
          if (t1-h1+>=) ans+=1ll*cnt*(cnt-)/*js(xx);
          ans+=1ll*ans2*cnt*(cnt-)/*(t1-h1+)*(t1-h1)/;
          break;
        }
        if (h-h1>=&&t1-t>=)
        {
          ans+=1ll*cnt*(cnt-)/*(h-h1)*(h-h1+)/*(t1-t)*(t1-t+)/;
        }
        ans+=1ll*cnt*(cnt-)/*ans2*(t1-t+)*(h-h1+);
        ans2+=(t1-t+)*(h-h1+);
        h++;
      }
    }
    if (cnt>=)
    {
      rg ll ans2=1ll*cnt*(cnt-)*(cnt-)/;
      unsigned rg ll ans4=;
      for (rg int i=;i<=n;i++)
      {
        if (a[now]>a[i])
        {
          int x=find1(a[now]-a[i]);
          int y=find2(i,a[now]-a[i]);
          ans4+=1ll*(x-y)*ans2;
        }
      }
      ans3+=ans4/;
    }
  }
  //ans3/=3;
  cout<<ans+ans3<<endl;
  return ;
}

对拍 n^6

#include <bits/stdc++.h>
using namespace std;
#define rg register
#define N 10000
#define rep(i,x,y) for (rg int i=x;i<=y;i++)
int b[];
bool cmp(int x,int y)
{
  return(x<y);
}
int a[N],n;
bool pd1(int x1,int x2,int x3,int x4,int x5,int x6)
{
  b[]=a[x1],b[]=a[x2],b[]=a[x3],b[]=a[x4],b[]=a[x5],b[]=a[x6];
  sort(b+,b++,cmp);
  if (b[]+b[]==b[]+b[]&&b[]+b[]==b[]&&b[]==b[])
    return(); else return();
}
bool pd2(int x1,int x2,int x3,int x4,int x5,int x6)
{
    b[]=a[x1],b[]=a[x2],b[]=a[x3],b[]=a[x4],b[]=a[x5],b[]=a[x6];
  sort(b+,b++,cmp);
  if (b[]+b[]+b[]==b[]&&b[]==b[]&&b[]==b[])
    return(); else return();
}
int main()
{
  freopen("noi.in","r",stdin);
  freopen("noi2.out","w",stdout);
  std::ios::sync_with_stdio(false);
  cin>>n;
  for (int i=;i<=n;i++) cin>>a[i];
  int cnt=,cnt2=;
  rep(i1,,n)
    rep(i2,i1+,n)
      rep(i3,i2+,n)
        rep(i4,i3+,n)
          rep(i5,i4+,n)
            rep(i6,i5+,n)
              if (pd1(i1,i2,i3,i4,i5,i6))
              {

                cnt++;
              }
  rep(i1,,n)
    rep(i2,i1+,n)
      rep(i3,i2+,n)
        rep(i4,i3+,n)
          rep(i5,i4+,n)
            rep(i6,i5+,n)
              if (pd2(i1,i2,i3,i4,i5,i6))
             {
          //      cout<<i1<<" "<<i2<<" "<<i3<<" "<<i4<<" "<<i5<<" "<<i6<<endl;
                cnt++;
              }
  cout<<cnt+cnt2<<endl;
  return ;
}