给定一个整数N ，任务是找到三个正整数A ， B和C ，使得表达式(3 * A + 5 * B + 7 * C)的值等于N。如果不存在这样的三元组，则打印“ -1” 。
例子：
Input: N = 19
Output:
A = 0
B = 1
C = 2
Explanation: Setting A, B, and C equal to 0, 1, and 2 respectively, the evaluated value of the expression = 3 * A + 5 * B + 7 * C = 3 * 0 + 5 * 1 + 7 * 2 = 19, which is the same as N (= 19).
Input: N = 4
Output: -1
天真的方法：解决问题的最简单方法是生成所有可能的三元组，其整数最大为N，并检查是否存在任何三元组(A，B，C)，以使值(3 * A + 5 * B + 7 * C)等于N。如果发现是真实的，则打印该三元组。否则，打印“ -1” 。
时间复杂度： O(N 3 )
辅助空间： O(1)
高效方法：可以基于以下观察结果来优化上述方法： A的值在[0，N / 3]范围内， B的值在[0，N / 5]范围内，并且值C的范围在[0，N / 7]范围内。请按照以下步骤解决问题：
- 迭代范围[0，N / 7]并执行以下操作：
- 在[0，N / 5]范围内迭代，并将A的值确定为(N – 5 * j – 7 * i) 。
- 在上述步骤中，如果A的值至少为0并且A可被3整除，则存在一个三元组，如(A / 3，i，j) 。打印此三元组并跳出循环。
- 完成上述步骤后，如果不存在任何此类三元组，请打印“ -1” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find a triplet (A, B, C)
// such that 3 * A + 5 * B + 7 * C is N
void CalculateValues(int N){
int A = 0, B = 0, C = 0;
// Iterate over the range [0, N//7]
for (C = 0; C < N/7; C++)
{
// Iterate over the range [0, N//5]
for ( B = 0; B < N/5; B++)
{
// Find the value of A
int A = N - 7 * C - 5 * B;
// If A is greater than or equal
// to 0 and divisible by 3
if (A >= 0 && A % 3 == 0)
{
cout << "A = " << A / 3 << ", B = " << B << ", C = "<< C << endl;
return;
}
}
}
// Otherwise, print -1
cout << -1 << endl;
}
// Driver Code
int main()
{
int N = 19;
CalculateValues(19);
return 0;
}
// This code is contributed by susmitakundugoaldanga. Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find a triplet (A, B, C)
// such that 3 * A + 5 * B + 7 * C is N
static void CalculateValues(int N)
{
int A = 0, B = 0, C = 0;
// Iterate over the range [0, N//7]
for (C = 0; C < N/7; C++)
{
// Iterate over the range [0, N//5]
for ( B = 0; B < N/5; B++)
{
// Find the value of A
A = N - 7 * C - 5 * B;
// If A is greater than or equal
// to 0 and divisible by 3
if (A >= 0 && A % 3 == 0)
{
System.out.print("A = " + A / 3 + ", B = " + B + ", C = "+ C);
return;
}
}
}
// Otherwise, print -1
System.out.println(-1);
}
// Driver Code
public static void main(String[] args)
{
int N = 19;
CalculateValues(19);
}
}
// This code is contributed by souravghosh0416.Python3
# Python program for the above approach
# Function to find a triplet (A, B, C)
# such that 3 * A + 5 * B + 7 * C is N
def CalculateValues(N):
# Iterate over the range [0, N//7]
for C in range(0, N//7 + 1):
# Iterate over the range [0, N//5]
for B in range(0, N//5 + 1):
# Find the value of A
A = N - 7 * C - 5 * B
# If A is greater than or equal
# to 0 and divisible by 3
if (A >= 0 and A % 3 == 0):
print("A =", A / 3, ", B =", B, ", \
C =", C, sep =" ")
return
# Otherwise, print -1
print(-1)
return
# Driver Code
if __name__ == '__main__':
N = 19
CalculateValues(19)C#
// C# program for the above approach
using System;
class GFG{
// Function to find a triplet (A, B, C)
// such that 3 * A + 5 * B + 7 * C is N
static void CalculateValues(int N)
{
int A = 0, B = 0, C = 0;
// Iterate over the range [0, N//7]
for (C = 0; C < N/7; C++)
{
// Iterate over the range [0, N//5]
for ( B = 0; B < N/5; B++)
{
// Find the value of A
A = N - 7 * C - 5 * B;
// If A is greater than or equal
// to 0 and divisible by 3
if (A >= 0 && A % 3 == 0)
{
Console.Write("A = " + A / 3 + ", B = " + B + ", C = "+ C);
return;
}
}
}
// Otherwise, print -1
Console.WriteLine(-1);
}
// Driver Code
static public void Main()
{
int N = 19;
CalculateValues(19);
}
}
// This code is contributed by splevel62.输出：
A = 3.0 , B = 2 , C = 0时间复杂度： O(N 2 )
辅助空间： O(1)