A * C大于B * B的三元组(A，B，C)的计数

给定三个整数A ， B和C。任务是计算三元组(a，b，c)的数量，以使a * c> b ² ，其中0 ， 0 和0
例子：

Input: A = 3, B = 2, C = 2 
Output: 6 
Following triples are counted : 
(1, 1, 2), (2, 1, 1), (2, 1, 2), (3, 1, 1), (3, 1, 2) and (3, 2, 2).
Input: A = 3, B = 3, C = 3 
Output: 11 
 

天真的方法：
蛮力方法是考虑所有可能的三元组(a，b，c)并计算满足约束a * c> b ^2的那些三元组。
下面是给定方法的实现。

C++

// C++ implementation
#include 
using namespace std;
 
// function to return the count
// of the valid triplets
long long countTriplets(int A, int B, int C)
{
    long long ans = 0;
    for (int i = 1; i <= A; i++) {
        for (int j = 1; j <= B; j++) {
            for (int k = 1; k <= C; k++) {
                if (i * k > j * j)
                    ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int A, B, C;
    A = 3, B = 2, C = 2;
 
    // function calling
    cout << countTriplets(A, B, C);
}

Java

// Java implementation of above approach
import java.util.*;
 
class GFG
{
 
// function to return the count
// of the valid triplets
static long countTriplets(int A, int B, int C)
{
    long ans = 0;
    for (int i = 1; i <= A; i++)
    {
        for (int j = 1; j <= B; j++)
        {
            for (int k = 1; k <= C; k++)
            {
                if (i * k > j * j)
                    ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
    int A = 3, B = 2, C = 2;
 
    // function calling
    System.out.println(countTriplets(A, B, C));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation for above approach
 
# function to return the count
# of the valid triplets
def countTriplets(A, B, C):
    ans = 0
    for i in range(1, A + 1):
        for j in range(1, B + 1):
            for k in range(1, C + 1):
                if (i * k > j * j):
                    ans += 1
 
    return ans
 
# Driver Code
A = 3
B = 2
C = 2
 
# function calling
print(countTriplets(A, B, C))
 
# This code is contributed by Mohit Kumar

C#

// C# implementation of above approach
using System;
 
class GFG
{
 
// function to return the count
// of the valid triplets
static long countTriplets(int A,
                          int B, int C)
{
    long ans = 0;
    for (int i = 1; i <= A; i++)
    {
        for (int j = 1; j <= B; j++)
        {
            for (int k = 1; k <= C; k++)
            {
                if (i * k > j * j)
                    ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
public static void Main (String[] args)
{
    int A = 3, B = 2, C = 2;
 
    // function calling
    Console.WriteLine(countTriplets(A, B, C));
}
}
     
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation
 
// function to return the count
// of the valid triplets
function countTriplets(A, B, C)
{
    let ans = 0;
    for (let i = 1; i <= A; i++) {
        for (let j = 1; j <= B; j++) {
            for (let k = 1; k <= C; k++) {
                if (i * k > j * j)
                    ans++;
            }
        }
    }
    return ans;
}
 
// Driver Code
    let A, B, C;
    A = 3, B = 2, C = 2;
 
    // function calling
    document.write(countTriplets(A, B, C));
 

输出：

6

时间复杂度： $O(A*B*C)$ 。
高效的方法：
让我们计算从1到B的所有k的给定值b = k的所有三胞胎。

1. 对于给定的b = k，我们需要找到满足i * j> k ^2的所有a = i和c = j
2. 对于a = i ，找到满足条件的最小c = j 。
   由于c = j满足此条件，因此c = j + 1，c = j + 2，… ，依此类推，也将满足该条件。
   因此，我们可以轻松地计算a = i和b = k的所有三元组。
3. 同样，如果对于某些a = i ， c = j是满足给定条件的最小值，则可以观察到a = j且所有c> = i也都满足该条件。
   该条件还通过a = j +1和c> = i来满足，即所有值a> = j和c> = i也都满足该条件。
4. 上面的观察有助于我们轻松地计算b = k和a> = j的所有三元组。
5. 现在我们需要计算b = k且i 所有三元组。
6. 因此，对于给定的b = k值，我们只需要向上求a = k的平方根。

下面是上述方法的实现：

C++

// C++ implementation
#include 
using namespace std;
 
// Counts the number of triplets
// for a given value of b
long long getCount(int A, int B2,
                   int C)
{
    long long count = 0;
 
    // Count all triples in which a = i
    for (int i = 1; i <= A; i++) {
 
        // Smallest value j
        // such that i*j > B2
        long long j = (B2 / i) + 1;
 
        // Count all (i, B2, x)
        // such that x >= j
        if (C >= j)
            count = (count + C - j + 1);
 
        // count all (x, B2, y) such
        // that x >= j this counts
        // all such triples in
        // which a >= j
        if (A >= j && C >= i)
            count = (count
                     + (C - i + 1)
                           * (A - j + 1));
 
        // As all triples with a >= j
        // have been counted reduce
        // A to j - 1.
        if (A >= j)
            A = j - 1;
    }
    return count;
}
 
// Counts the number of triples that
// satisfy the given constraints
long long countTriplets(int A, int B,
                        int C)
{
    long long ans = 0;
    for (int i = 1; i <= B; i++) {
 
        // GetCount of triples in which b = i
        ans = (ans
               + getCount(A, i * i, C));
    }
    return ans;
}
 
// Driver Code
int main()
{
    int A, B, C;
    A = 3, B = 2, C = 2;
 
    // Function calling
    cout << countTriplets(A, B, C);
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Counts the number of triplets
// for a given value of b
static long getCount(int A, int B2, int C)
{
    long count = 0;
 
    // Count all triples in which a = i
    for (int i = 1; i <= A; i++)
    {
 
        // Smallest value j
        // such that i*j > B2
        long j = (B2 / i) + 1;
 
        // Count all (i, B2, x)
        // such that x >= j
        if (C >= j)
            count = (count + C - j + 1);
 
        // count all (x, B2, y) such
        // that x >= j this counts
        // all such triples in
        // which a >= j
        if (A >= j && C >= i)
            count = (count + (C - i + 1) *
                             (A - j + 1));
 
        // As all triples with a >= j
        // have been counted reduce
        // A to j - 1.
        if (A >= j)
            A = (int) (j - 1);
    }
    return count;
}
 
// Counts the number of triples that
// satisfy the given constraints
static long countTriplets(int A, int B, int C)
{
    long ans = 0;
    for (int i = 1; i <= B; i++)
    {
 
        // GetCount of triples in which b = i
        ans = (ans + getCount(A, i * i, C));
    }
    return ans;
}
 
// Driver Code
public static void main(String[] args)
{
    int A, B, C;
    A = 3; B = 2; C = 2;
 
    // Function calling
    System.out.println(countTriplets(A, B, C));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 implementation
 
# Counts the number of triplets
# for a given value of b
def getCount(A, B2, C):
     
    count = 0
     
    # Count all triples in which a = i
    i=1
    while(i B2
        j = (B2 // i) + 1
        # Count all (i, B2, x)
        # such that x >= j
        if (C >= j):
            count = count + C - j + 1
             
        # count all (x, B2, y) such
        # that x >= j this counts
        # all such triples in
        # which a >= j
        if (A>= j and C >= i):
            count = count+ (C - i + 1)    * (A - j + 1)
             
        # As all triples with a >= j
        # have been counted reduce
        # A to j - 1.
        if (A >= j):
            A = j - 1
        i+=1
     
    return count
 
 
# Counts the number of triples that
# satisfy the given constraints
def countTriplets(A, B, C):
     
    ans = 0
    for i in range(1,B+1):
        # GetCount of triples in which b = i
        ans = (ans+ getCount(A, i * i, C))
     
    return ans
 
 
# Driver Code
 
A = 3
B = 2
C = 2
 
# Function calling
print(countTriplets(A, B, C))
 
# This code is contributed by shubhamsingh10

C＃

// C# implementation of the approach
using System;
using System.Collections.Generic;                
     
class GFG
{
 
// Counts the number of triplets
// for a given value of b
static long getCount(int A, int B2, int C)
{
    long count = 0;
 
    // Count all triples in which a = i
    for (int i = 1; i <= A; i++)
    {
 
        // Smallest value j
        // such that i*j > B2
        long j = (B2 / i) + 1;
 
        // Count all (i, B2, x)
        // such that x >= j
        if (C >= j)
            count = (count + C - j + 1);
 
        // count all (x, B2, y) such
        // that x >= j this counts
        // all such triples in
        // which a >= j
        if (A >= j && C >= i)
            count = (count + (C - i + 1) *
                             (A - j + 1));
 
        // As all triples with a >= j
        // have been counted reduce
        // A to j - 1.
        if (A >= j)
            A = (int) (j - 1);
    }
    return count;
}
 
// Counts the number of triples that
// satisfy the given constraints
static long countTriplets(int A, int B, int C)
{
    long ans = 0;
    for (int i = 1; i <= B; i++)
    {
 
        // GetCount of triples in which b = i
        ans = (ans + getCount(A, i * i, C));
    }
    return ans;
}
 
// Driver Code
public static void Main(String[] args)
{
    int A, B, C;
    A = 3; B = 2; C = 2;
 
    // Function calling
    Console.WriteLine(countTriplets(A, B, C));
}
}
 
// This code is contributed by Princi Singh

Java脚本

 
// Javascript implementation
 
// Counts the number of triplets
// for a given value of b
function getCount(A, B2, C)
{
    let count = 0;
 
    // Count all triples in which a = i
    for (let i = 1; i <= A; i++) {
 
        // Smallest value j
        // such that i*j > B2
        let j = parseInt(B2 / i) + 1;
 
        // Count all (i, B2, x)
        // such that x >= j
        if (C >= j)
            count = (count + C - j + 1);
 
        // count all (x, B2, y) such
        // that x >= j this counts
        // all such triples in
        // which a >= j
        if (A >= j && C >= i)
            count = (count
                     + (C - i + 1)
                           * (A - j + 1));
 
        // As all triples with a >= j
        // have been counted reduce
        // A to j - 1.
        if (A >= j)
            A = j - 1;
    }
    return count;
}
 
// Counts the number of triples that
// satisfy the given constraints
function countTriplets(A, B, )
{
    let ans = 0;
    for (let i = 1; i <= B; i++) {
 
        // GetCount of triples in which b = i
        ans = (ans
               + getCount(A, i * i, C));
    }
    return ans;
}
 
// Driver Code
    let A, B, C;
    A = 3, B = 2, C = 2;
 
    // Function calling
    document.write(countTriplets(A, B, C));
 

输出：

6

时间复杂度： $O(B^{2})$

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