使用给定的总和和平均值来查找缺失天的温度

鉴于整数x和y表示本周除了分别天第一天与第2天的平均温度，和第一天与第2天为S的温度的总和，任务是找到天第一天与第2天的温度。
例子：

Input: x = 15, y = 10, S = 50 
Output: Day1 = 10, Day2 = 40 
Explanation: 
The average of week excluding Day1 is 15, the average of week excluding Day2 is 10 and the sum of temperature of Day1 and Day2 is 50. Individual temperature of the two days are 10 and 40 respecticvely.
Input: x = 5, y = 10, s = 40 
Output: Day1 = 35, Day2 = 5 
Explanation: 
The average of week excluding Day1 is 5, the average of week excluding Day2 is 10 and the sum of temperature of Day1 and Day2 is 40. Individual temperature of the two days are 35 and 5 respecticvely. 
 

方法：我们知道平均值=所有观察值的总和/观察总数。因此，观测值的总和=平均值*观测值的数量，即S = A * n

So after excluding Day1 or Day2 we are left with only 6 days 
so N = 6 and the equations are: 
 

$total - Day1 = 6 * x$
 

and 
 

$total - Day2 = 6 * y$
 

on subtracting the above two equations we get, 
 

$Day2 - Day1 = ( x - y ) * 6$
 

(Equation 1) 
and it is given in the problem statement that 
 

$Day2 + Day1 = S$
 

(Equation 2) 
 

 

解决上述两个方程式，Day1和Day2的值由下式给出：

$Day2 = \frac{(S +(x-y)*6)}{2}$[Tex]Day1 = S – Day2  [/Tex]

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function for finding the temperature
void findTemperature(int x, int y, int s)
{
    double Day1, Day2;
 
    // Store Day1 - Day2 in diff
    double diff = (x - y) * 6;
 
    Day2 = (diff + s) / 2;
 
    // Remaining from s will be Day1
    Day1 = s - Day2;
 
    // Print Day1 and Day2
    cout << "Day1 : " << Day1 << endl;
    cout << "Day2 : " << Day2 << endl;
}
 
// Driver Code
int main()
{
    int x = 5, y = 10, s = 40;
 
    // Functions
    findTemperature(x, y, s);
 
    return 0;
}

Java

// Java program for the above approach
class GFG{
  
// Function for finding the temperature
static void findTemperature(int x, int y, int s)
{
    double Day1, Day2;
  
    // Store Day1 - Day2 in diff
    double diff = (x - y) * 6;
  
    Day2 = (diff + s) / 2;
  
    // Remaining from s will be Day1
    Day1 = s - Day2;
  
    // Print Day1 and Day2
    System.out.println( "Day1 : " + Day1);
    System.out.println( "Day2 : " + Day2);
}
  
// Driver Code
public static void main(String[] args)
{
    int x = 5, y = 10, s = 40;
  
    // Functions
    findTemperature(x, y, s);
}
}
 
// This code is contributed by rock_cool

Python

# Python3 program for the above approach
 
# Function for finding the temperature
def findTemperature(x, y, s):
 
    # Store Day1 - Day2 in diff
    diff = (x - y) * 6
    Day2 = (diff + s) // 2
 
    # Remaining from s will be Day1
    Day1 = s - Day2
 
    # Print Day1 and Day2
    print("Day1 : ", Day1)
    print("Day2 : ", Day2)
 
# Driver Code
if __name__ == '__main__':
    x = 5
    y = 10
    s = 40
 
    # Functions
    findTemperature(x, y, s)
 
# This code is contributed by Mohit Kumar

C#

// C# program for the above approach
using System;
class GFG{
   
// Function for finding the temperature
static void findTemperature(int x, int y, int s)
{
    double Day1, Day2;
   
    // Store Day1 - Day2 in diff
    double diff = (x - y) * 6;
   
    Day2 = (diff + s) / 2;
   
    // Remaining from s will be Day1
    Day1 = s - Day2;
   
    // Print Day1 and Day2
    Console.Write( "Day1 : " + Day1 + '\n');
    Console.WriteLine( "Day2 : " + Day2 + '\n');
}
   
// Driver Code
public static void Main(string[] args)
{
    int x = 5, y = 10, s = 40;
   
    // Functions
    findTemperature(x, y, s);
}
}
 
// This code is contributed by Ritik Bansal

Javascript

 
    // Javascript program for the above approach
     
    // Function for finding the temperature
    function findTemperature(x, y, s)
    {
        let Day1, Day2;
 
        // Store Day1 - Day2 in diff
        let diff = (x - y) * 6;
 
        Day2 = (diff + s) / 2;
 
        // Remaining from s will be Day1
        Day1 = s - Day2;
 
        // Print Day1 and Day2
        document.write("Day1 : " + Day1 + "</br>");
        document.write("Day2 : " + Day2 + "</br>");
    }
 
    let x = 5, y = 10, s = 40;
   
    // Functions
    findTemperature(x, y, s);
     

输出：

Day1 : 35
Day2 : 5

时间复杂度： O(1)
辅助空间： O(1)