给定一个由n个整数组成的数组，对所有三元组的总和等于理想立方的总数进行计数，即对于满足a [i] + a [j] + a的任何i，j，k(i
例子：
Input:
N = 5
2 5 1 20 6
Output:
3
Explanation:
There are only 3 triplets whose total sum is a perfect cube.
Indices Values SUM
0 1 2 2 5 1 8
0 1 3 2 5 20 27
2 3 4 1 20 6 27
Since 8 and 27 are prefect cube of 2 and 3.
天真的方法是使用3个嵌套循环迭代所有可能的数字，并检查它们的和是否为理想的立方体。由于时间复杂度可能高达O(n 3 )，因此该方法将非常慢。
一种有效的方法是使用动态编程和基础数学。根据给定条件，三个正整数之和不大于15000。因此，在1到15000范围内，只能有24(15000 1/3 )个立方体。
现在，无需遍历所有三胞胎，我们就可以借助上述信息来做得更好。固定了前两个索引i和j，这样我们就不必遍历所有k(j
但是，如果我们计算出现在a [j + 1，j + 2，…n]中的数字，例如K，那么这将再次被视为慢速方法，并且肯定会给出TLE。因此，我们必须考虑一种不同的方法。
现在是动态编程。由于所有数字均小于5000，n最多为1000。因此我们可以将DP数组计算为：
dp [i] [K]：= A [i，i + 1，i + 2…n]中K的出现次数
C++
// C++ program to calculate all triplets whose
// sum is perfect cube.
#include
using namespace std;
int dp[1001][15001];
// Function to calculate all occurrence of
// a number in a given range
void computeDpArray(int arr[], int n)
{
for (int i = 0; i < n; ++i) {
for (int j = 1; j <= 15000; ++j) {
// if i == 0
// assign 1 to present state
if (i == 0)
dp[i][j] = (j == arr[i]);
// else add +1 to current state with
// previous state
else
dp[i][j] = dp[i - 1][j] + (arr[i] == j);
}
}
}
// Function to calculate triplets whose sum
// is equal to the pefect cube
int countTripletSum(int arr[], int n)
{
computeDpArray(arr, n);
int ans = 0; // Initialize answer
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
for (int k = 1; k <= 24; ++k) {
int cube = k * k * k;
int rem = cube - (arr[i] + arr[j]);
// count all occurrence of third triplet
// in range from j+1 to n
if (rem > 0)
ans += dp[n - 1][rem] - dp[j][rem];
}
}
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 2, 5, 1, 20, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << countTripletSum(arr, n);
return 0;
} Java
// JAVA Code for Count all triplets whose
// sum is equal to a perfect cube
import java.util.*;
class GFG {
public static int dp[][];
// Function to calculate all occurrence of
// a number in a given range
public static void computeDpArray(int arr[], int n)
{
for (int i = 0; i < n; ++i) {
for (int j = 1; j <= 15000; ++j) {
// if i == 0
// assign 1 to present state
if (i == 0 && j == arr[i])
dp[i][j] = 1;
else if(i==0)
dp[i][j] = 0;
// else add +1 to current state
// with previous state
else if(arr[i] == j)
dp[i][j] = dp[i - 1][j] + 1;
else
dp[i][j] = dp[i - 1][j];
}
}
}
// Function to calculate triplets whose sum
// is equal to the pefect cube
public static int countTripletSum(int arr[], int n)
{
computeDpArray(arr, n);
int ans = 0; // Initialize answer
for (int i = 0; i < n - 2; ++i) {
for (int j = i + 1; j < n - 1; ++j) {
for (int k = 1; k <= 24; ++k) {
int cube = k * k * k;
int rem = cube - (arr[i] + arr[j]);
// count all occurrence of
// third triplet in range
// from j+1 to n
if (rem > 0)
ans += dp[n - 1][rem] - dp[j][rem];
}
}
}
return ans;
}
/* Driver program to test above function */
public static void main(String[] args)
{
int arr[] = { 2, 5, 1, 20, 6 };
int n = arr.length;
dp = new int[1001][15001];
System.out.println(countTripletSum(arr, n));
}
}
// This code is contributed by Arnav Kr. Mandal.Python3
# Python 3 program to calculate all
# triplets whose sum is perfect cube.
dp = [[0 for i in range(15001)]
for j in range(1001)]
# Function to calculate all occurrence
# of a number in a given range
def computeDpArray(arr, n):
for i in range(n):
for j in range(1, 15001, 1):
# if i == 0
# assign 1 to present state
if (i == 0):
dp[i][j] = (j == arr[i])
# else add +1 to current state with
# previous state
else:
dp[i][j] = dp[i - 1][j] + (arr[i] == j)
# Function to calculate triplets whose
# sum is equal to the pefect cube
def countTripletSum(arr, n):
computeDpArray(arr, n)
ans = 0 # Initialize answer
for i in range(0, n - 2, 1):
for j in range(i + 1, n - 1, 1):
for k in range(1, 25, 1):
cube = k * k * k
rem = cube - (arr[i] + arr[j])
# count all occurrence of third
# triplet in range from j+1 to n
if (rem > 0):
ans += dp[n - 1][rem] - dp[j][rem]
return ans
# Driver code
if __name__ == '__main__':
arr = [2, 5, 1, 20, 6]
n = len(arr)
print(countTripletSum(arr, n))
# This code is contributed by
# Sahil_ShelangiaC#
// C# Code for Count all triplets whose
// sum is equal to a perfect cube
using System;
class GFG
{
public static int [,]dp;
// Function to calculate all occurrence
// of a number in a given range
public static void computeDpArray(int []arr,
int n)
{
for (int i = 0; i < n; ++i)
{
for (int j = 1; j <= 15000; ++j)
{
// if i == 0
// assign 1 to present state
if (i == 0 && j == arr[i])
dp[i, j] = 1;
else if(i == 0)
dp[i, j] = 0;
// else add +1 to current state
// with previous state
else if(arr[i] == j)
dp[i, j] = dp[i - 1, j] + 1;
else
dp[i, j] = dp[i - 1, j];
}
}
}
// Function to calculate triplets whose
// sum is equal to the pefect cube
public static int countTripletSum(int []arr,
int n)
{
computeDpArray(arr, n);
int ans = 0; // Initialize answer
for (int i = 0; i < n - 2; ++i)
{
for (int j = i + 1; j < n - 1; ++j)
{
for (int k = 1; k <= 24; ++k)
{
int cube = k * k * k;
int rem = cube - (arr[i] + arr[j]);
// count all occurrence of
// third triplet in range
// from j+1 to n
if (rem > 0)
ans += dp[n - 1, rem] -
dp[j, rem];
}
}
}
return ans;
}
// Driver Code
public static void Main()
{
int []arr = { 2, 5, 1, 20, 6 };
int n = arr.Length;
dp = new int[1001, 15001];
Console.Write(countTripletSum(arr, n));
}
}
// This code is contributed
// by 29AjayKumarPHP
0)
$ans += $dp[$n - 1][$rem] -
$dp[$j][$rem];
}
}
}
return $ans;
}
// Driver code
$arr = array(2, 5, 1, 20, 6);
$n = sizeof($arr);
echo countTripletSum($arr, $n);
// This code is contributed by ita_c
?>输出：
3
时间复杂度： O(N 2 * 24)
辅助空间： O(10 7 )