Home  >  Article  >  Backend Development  >  Using C++ programming, find the number of solutions to the equation x + y + z <= n

Using C++ programming, find the number of solutions to the equation x + y + z <= n

王林
王林forward
2023-08-25 11:41:241482browse

使用C++编程,找到方程x + y + z <= n的解的数量

In this article, we will explain the method on how to find the solution of the equation x y z<=n. In this problem, we have an equation with four variables and the task is to find the solution of the given equation. So here is a simple example&miuns;

Input: X = 1, Y = 1, Z = 1, n = 1

Output: 4

Input: X = 1, Y = 2, Z = 3, n = 4

Output: 3

In this problem we can simply loop through all (x, y), (y,z) by isolating each variable and checking if it satisfies the equation , the value of (x,z).

Solution

Now we will use brute force method to find the solution to the given problem.

Brute force method

In this program, we will traverse all possible values ​​of (x,y), (y,z) and (x,z) to satisfy the equation z

Example


#include<bits/stdc++.h>
using namespace std;
int main(){
    int X = 1, Y = 2, Z = 3, n = 4; // limits of x, y, z and given n.
    int answer = 0; // counter variable.
    for(int i = 0; i <= X; i++){
        for(int j = 0; j <= Y; j++){
            int temp = (n - i) - j; // temp = n - x - y.
            if(temp >= Z){ // if n - x - y >= z so we increment the answer.
               answer++;
            }
        }
    }
    for(int i = 0; i <= X; i++){
        for(int j = 0; j <= Z; j++){
            int temp = (n - i) - j; // temp = n - x - y.
            if(temp >= Y){ // if n - x - y >= z so we increment the answer.
               answer++;
            }
        }
    }
    for(int i = 0; i <= Z; i++){
        for(int j = 0; j <= Y; j++){
            int temp = (n - i) - j; // temp = n - x - y.
            if(temp >= X){ // if n - x - y >= z so we increment the answer.
               answer++;
            }
        }
    }
    cout << answer << "\n";
}

Output

17

Explanation of the above program

In this program, we will use a nested for loop to traverse all (x ,y), (y, z) and (x, z), and check whether the equation satisfies the condition, and if so, add the answer.

Conclusion

In this article, we solved a problem to find the number of solutions that satisfy the equation x y z<= n, with a time complexity of O(X*Y). We also learned a C program to solve this problem and the complete solution. We can write the same program in other languages ​​like C, Java, Python and others.

The above is the detailed content of Using C++ programming, find the number of solutions to the equation x + y + z <= n. For more information, please follow other related articles on the PHP Chinese website!

Statement:
This article is reproduced at:tutorialspoint.com. If there is any infringement, please contact admin@php.cn delete