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📜  检查在给定的操作下是否有可能使数组的和为奇数

📅  最后修改于: 2021-05-17 22:00:18             🧑  作者: Mango

给定一个数组arr [] ,任务是检查是否有可能使数组的和成为奇数,以便可以选择任意两个索引ij,并且在给定arr [i]的情况下可以将arr [i]设置为等于arr [j]我!= j

例子:

方法:想法是找到数组中所有元素的总和,并同时检查数组中存在的偶数和奇数。如果总和为偶数,则可以用奇数代替偶数之一,从而使数组的总和为奇数。如果数组中没有奇数元素,则数组的总和不能为奇数。

下面是上述方法的实现:

C++
// C++ program to check if the array
// with odd sum is possible
#include 
using namespace std;
  
// Function to check if the
// sum of the array can be made odd.
bool isOdd(int arr[], int n)
{
    int l, r, flag = 0, flag1 = 0, sum = 0;
  
    // Find sum of all elements and increment
    // check for odd or even elements in the array
    // so that by changing ai=aj,
    // the sum of the array can be made odd
    for (int i = 0; i < n; i++) {
        sum += arr[i];
        if (arr[i] % 2 == 0 && flag == 0) {
            flag = 1;
            l = arr[i];
        }
        if (arr[i] % 2 != 0 && flag1 == 0) {
            r = arr[i];
            flag1 = 1;
        }
    }
  
    // If the sum is already odd
    if (sum % 2 != 0) {
        return true;
    }
  
    // Else, then both the flags should be checked.
    // Here, flag1 and flag represent if there is
    // an odd-even pair which can be replaced.
    else {
        if (flag1 == 1 && flag == 1)
            return true;
        else
            return false;
    }
}
  
// Driver code
int main()
{
    int ar[] = { 5, 4, 4, 5, 1, 3 };
    int n = sizeof(ar) / sizeof(ar[0]);
    bool res = isOdd(ar, n);
  
    if (res)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}


Java
// Java program to check if the array
// with odd sum is possible
class GFG {
      
    // Function to check if the
    // sum of the array can be made odd.
    static boolean isOdd(int []arr, int n)
    {
        int l, r, flag = 0, flag1 = 0, sum = 0;
      
        // Find sum of all elements and increment
        // check for odd or even elements in the array
        // so that by changing ai=aj,
        // the sum of the array can be made odd
        for (int i = 0; i < n; i++) {
            sum += arr[i];
            if (arr[i] % 2 == 0 && flag == 0) {
                flag = 1;
                l = arr[i];
            }
            if (arr[i] % 2 != 0 && flag1 == 0) {
                r = arr[i];
                flag1 = 1;
            }
        }
      
        // If the sum is already odd
        if (sum % 2 != 0) {
            return true;
        }
      
        // Else, then both the flags should be checked.
        // Here, flag1 and flag represent if there is
        // an odd-even pair which can be replaced.
        else {
            if (flag1 == 1 && flag == 1)
                return true;
            else
                return false;
        }
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int ar[] = { 5, 4, 4, 5, 1, 3 };
        int n = ar.length;
        boolean res = isOdd(ar, n);
      
        if (res == true)
            System.out.println("Yes");
        else
            System.out.println("No");
  
    }
}
  
// This code is contributed by AnkitRai01


Python3
# Python3 program to check if the array
# with odd sum is possible
  
# Function to check if the
# sum of the array can be made odd.
def isOdd(arr,  n) :
    flag = 0; flag1 = 0; sum = 0;
  
    # Find sum of all elements and increment
    # check for odd or even elements in the array
    # so that by changing ai=aj,
    # the sum of the array can be made odd
    for i in range(n) :
        sum += arr[i];
        if (arr[i] % 2 == 0 and flag == 0) :
            flag = 1;
            l = arr[i];
          
        if (arr[i] % 2 != 0 and flag1 == 0) :
            r = arr[i];
            flag1 = 1;
  
    # If the sum is already odd
    if (sum % 2 != 0) :
        return True;
  
    # Else, then both the flags should be checked.
    # Here, flag1 and flag represent if there is
    # an odd-even pair which can be replaced.
    else :
        if (flag1 == 1 and flag == 1) :
            return True;
        else :
            return False;
  
# Driver code
if __name__ == "__main__" :
  
    arr = [ 5, 4, 4, 5, 1, 3 ];
    n = len(arr);
      
    res = isOdd(arr, n);
  
    if (res) :
        print("Yes");
    else :
        print("No");
     
# This code is contributed by AnkitRai01


C#
// C# program to check if the array
// with odd sum is possible
using System;
  
class GFG{
    // Function to check if the
    // sum of the array can be made odd.
    static bool isOdd(int[] arr, int n)
    {
        int flag = 0, flag1 = 0, sum = 0;
      
        // Find sum of all elements and increment
        // check for odd or even elements in the array
        // so that by changing ai=aj,
        // the sum of the array can be made odd
        for (int i = 0; i < n; i++) {
            sum += arr[i];
            if (arr[i] % 2 == 0 && flag == 0) {
                flag = 1;
            }
            if (arr[i] % 2 != 0 && flag1 == 0) {
                flag1 = 1;
            }
        }
      
        // If the sum is already odd
        if (sum % 2 != 0) {
            return true;
        }
      
        // Else, then both the flags should be checked.
        // Here, flag1 and flag represent if there is
        // an odd-even pair which can be replaced.
        else {
            if (flag1 == 1 && flag == 1)
                return true;
            else
                return false;
        }
    }
      
    // Driver code   
    static public void Main ()
    {
        int[] ar = { 5, 4, 4, 5, 1, 3 };
        int n = ar.Length;
        bool res = isOdd(ar, n);
      
        if (res)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
  
// This code is contributed by shivanisingh


输出:
Yes

时间复杂度: O(N)