仅通过加或减给定的秒来检查是否有可能回到12’0时钟

给定N秒。任务是检查是否有可能从12’0时钟开始并仅增加或减去给定的秒数回到12。我们需要只使用给定的所有秒数一次，我们可以添加元素也可以减去元素。

例子：

Input: a[] = {60, 60, 120} 
Output: YES 
Add the first two seconds and 
subtract the last one to get back to 0. 

Input : a[] = {10, 20, 60, 180} 
Output : NO 

简单方法：生成所有可能的组合以解决上述问题。因此，生成N个数的幂集。检查任何人的总和％(24 * 60)是否等于零，如果等于零，则有可能不等于零。

下面是上述方法的实现：

C++

// C++ program to check if we come back to
// zero or not in a clock
#include 
using namespace std;
  
// Function to check all combinations
bool checkCombinations(int a[], int n)
{
  
    // Generate all power sets
    int pow_set_size = pow(2, n);
    int counter, j;
  
    // Check for every combination
    for (counter = 0; counter < pow_set_size; counter++) {
  
        // Store sum for all combiantions
        int sum = 0;
        for (j = 0; j < n; j++) {
  
            /* Check if jth bit in the counter is set 
             If set then print jth element from set */
            if (counter & (1 << j))
                sum += a[j]; // if set then consider as '+'
            else
                sum -= a[j]; // else consider as '-'
        }
  
        // If we can get back to 0
        if (sum % (24 * 60) == 0)
            return true;
    }
    return false;
}
// Driver Code
int main()
{
    int a[] = { 60, 60, 120 };
    int n = sizeof(a) / sizeof(a[0]);
  
    if (checkCombinations(a, n))
        cout << "YES";
    else
        cout << "NO";
    return 0;
}

Java

// Java program to check if we come 
// back to zero or not in a clock
import java.lang.Math;
  
class GfG
{
  
    // Function to check all combinations 
    static boolean checkCombinations(int a[], int n) 
    { 
        // Generate all power sets 
        int pow_set_size = (int)Math.pow(2, n); 
        int counter, j; 
      
        // Check for every combination 
        for (counter = 0; counter < pow_set_size; counter++) 
        { 
      
            // Store sum for all combiantions 
            int sum = 0; 
            for (j = 0; j < n; j++) 
            { 
      
                /* Check if jth bit in the counter is set 
                If set then print jth element from set */
                if ((counter & (1 << j)) != 0) 
                    sum += a[j]; // if set then consider as '+' 
                else
                    sum -= a[j]; // else consider as '-' 
            } 
      
            // If we can get back to 0 
            if (sum % (24 * 60) == 0) 
                return true; 
        }
        return false; 
    } 
  
    // Driver code
    public static void main(String []args)
    {
        int a[] = { 60, 60, 120 }; 
        int n = a.length; 
      
        if (checkCombinations(a, n)) 
            System.out.println("YES"); 
        else
            System.out.println("NO");
    }
}
      
// This code is contributed by Rituraj Jain

Python 3

# Python 3 program to check if we come 
# back to zero or not in a clock
  
# Function to check all combinations
def checkCombinations(a, n):
  
    # Generate all power sets
    pow_set_size = pow(2, n)
  
    # Check for every combination
    for counter in range(pow_set_size):
  
        # Store sum for all combiantions
        sum = 0
        for j in range(n) :
  
            # Check if jth bit in the counter is set 
            # If set then print jth element from set 
            if (counter & (1 << j)):
                sum += a[j] # if set then consider as '+'
            else:
                sum -= a[j] # else consider as '-'
  
        # If we can get back to 0
        if (sum % (24 * 60) == 0):
            return True
    return False
  
# Driver Code
if __name__ == "__main__":
      
    a = [ 60, 60, 120 ]
    n = len(a)
  
    if (checkCombinations(a, n)):
        print("YES")
    else:
        print("NO")
  
# This code is contributed by ita_c

C#

// C# program to check if we come 
// back to zero or not in a clock
using System;
  
class GfG
{
  
    // Function to check all combinations 
    static bool checkCombinations(int [] a, int n) 
    { 
        // Generate all power sets 
        int pow_set_size = (int)Math.Pow(2, n); 
        int counter, j; 
      
        // Check for every combination 
        for (counter = 0; counter < pow_set_size; counter++) 
        { 
      
            // Store sum for all combiantions 
            int sum = 0; 
            for (j = 0; j < n; j++) 
            { 
      
                /* Check if jth bit in the counter is set 
                If set then print jth element from set */
                if ((counter & (1 << j)) != 0) 
                    sum += a[j]; // if set then consider as '+' 
                else
                    sum -= a[j]; // else consider as '-' 
            } 
      
            // If we can get back to 0 
            if (sum % (24 * 60) == 0) 
                return true; 
        }
        return false; 
    } 
  
    // Driver code
    public static void Main()
    {
        int [] a = { 60, 60, 120 }; 
        int n = a.Length; 
      
        if (checkCombinations(a, n)) 
            Console.WriteLine("YES"); 
        else
            Console.WriteLine("NO");
    }
}
      
// This code is contributed by ihritik

输出：

YES

如果仔细研究，我们会发现该问题基本上是分区问题的一种变体。因此，我们可以使用动态编程对其进行优化(请参阅分区问题的方法2)。

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