3长度AP的阵列最快的算法计数 [英] fastest algorithm count number of 3 length AP in array

问题描述

我要解决这 codeChef挑战：

I want to solve this CodeChef challenge:

假设我们给出N的数组A（范围100,000）的元素。我们要找出所有对3个这样的元素1 LT的数量=爱，AJ，阿克＆LT; = 30,000使得

  AJ-艾= AK-AJ和I＆LT; J＆LT;氏/ code>
 
换句话说艾，AJ，阿克都在算术级数。例如，对于阵：


Suppose We are given an array A of N(of range 100,000) elements. We are to find the count of all pairs of 3 such elements 1<=Ai,Aj,Ak<=30,000 such that

Aj-Ai = Ak- Aj and i < j < k

In other words Ai,Aj,Ak are in Arithmetic Progression. For instance for Array :

9 4 2 3 6 10 3 3 10

所以AP是：

so The AP are:

{2,6,10},{9,6,3},{9,6,3},{3,3,3},{2,6,10} 

因此，所需的答案是5。

So the required answer is 5.

我试过是拿3万评为过去，右长的阵列。最初，右包含每个1-30,000元素的数量。

如果我们在第i个位置经过许多商店数组值的前阵我和右侧存储计计数后，我。我简单地循环为阵列中的所有可能的公差。这里是code：

What I tried is take 30,000 long arrays named past and right. Initially right contains the count of each 1-30,000 element.

If we are at ith position past stores the count of array value before i and right stores the count of array after i. I simply loop for all possible common difference in the array. Here is the code :

right[arr[1]]--;

for(i=2;i<=n-1;i++)
{
    past[arr[i-1]]++;
    right[arr[i]]--;
    k=30000 - arr[i];
    if(arr[i] <= 15000)
        k=arr[i];
    for(d=1;d<=k;d++)
    {
        ans+= right[arr[i] + d]*past[arr[i]-d] + past[arr[i] + d]*right[arr[i]-d];
    }
    ans+=past[arr[i]]*right[arr[i]];
}

但是，这让我的时间超出限制。请帮助一个更好的算法。

But this gets me Time Limit Exceeded. Please help with a better algorithm.

推荐答案

您可以大大节省运行时间，如果你犯了一个先传过来的名单，只提取数对，有可能有（差之间的3学期AP是0模2）。而这种对之间再进行迭代。

You can greatly cut execution time if you make a first pass over the list and only extract number pairs that it is possible to have an 3 term AP between (difference is 0 mod 2). And then iterating between such pairs.

伪C ++ - Ÿcode：

Pseudo C++-y code:

// Contains information about each beginning point
struct BeginNode {
  int value;
  size_t offset;
  SortedList<EndNode> ends;  //sorted by EndNode.value
};

// Contains information about each class of end point
struct EndNode {
  int value;
  List<size_t> offsets; // will be sorted without effort due to how we collect offsets
};

struct Result {
  size_t begin;
  size_t middle;
  size_t end;
};

SortedList<BeginNode> nodeList;
foreach (auto i : baseList) {
  BeginNode begin;
  node.value = i;
  node.offset = i's offset; //you'll need to use old school for (i=0;etc;i++) with this
  // baseList is the list between begin and end-2 (inclusive)
  foreach (auto j : restList) { 
    // restList is the list between iterator i+2 and end (inclusive)
    // we do not need to consider i+1, because not enough space for AP
    if ((i-j)%2 == 0) { //if it's possible to have a 3 term AP between these two nodes
      size_t listOffset = binarySearch(begin.ends);
      if (listOffset is valid) {
        begin.ends[listOffset].offsets.push_back(offsets);
      } else {
        EndNode end;
        end.value = j;
        end.offsets.push_back(j's offset);
        begin.ends.sorted_insert(end);
      }
    }
  }
  if (begin has shit in it) {
    nodeList.sorted_insert(begin);
  }
}
// Collection done, now iterate over collection

List<Result> res;
foreach (auto node : nodeList) {
  foreach (auto endNode : node.ends) {
    foreach (value : sublist from node.offset until endNode.offsets.last()) {
      if (value == average(node.value, endNode.value)) {
        // binary_search here to determine how many offsets in "endNode.offsets" "value's offset" is less than.
        do this that many times:
          res.push_back({node.value, value, endNode.value});
      }
    }
  }
}

return res;

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