在本文中,我們將找到方程式n = x n ⊕ x 的解的數量,即我們需要找到給定值n 的可能的x 值的數量,使得n = x n ⊕ x,其中⊕ 表示異或操作。
現在我們將討論關於 n = x n ⊕ x 的解的數量的完整信息,並提供適當的示例。
我們可以簡單地使用暴力法來找到解的數量,即對於給定的n 值,我們從0 開始應用每個整數值的x,並驗證方程是否滿足,x 的值應小於或等於n,因為將大於n 的值與(n ⊕ x) 相加將永遠不會傳回n 作為答案。
找到一個使 n = 3 成立的 x 的值?
n = x + n ⊕ x Putting x = 0, 3 = 0 + 3 ⊕ 0 3 ⊕ 0 = 3, 3 = 3 LHS = RHS(x = 0 satisfy the equation) So, x = 0 is one of the solution
#include <bits/stdc++.h> using namespace std; int main(){ int n = 3, c=0; for (int x = 0; x <= n; ++x)// loop for giving value of x from 0 to n if (n == x + n ^ x)//checking if value of x satisfies the equation ++c; cout << "Number of possible solutions : " << c; return 0; }
Number of possible solutions : 4
這是一個簡單的C 程序,透過應用暴力方法來找到n = x n ⊕ x的解的數量。
在這個方法中,如果我們看二進位形式的n,我們需要找到設定為1的位數,並且根據方程,我們可以說如果n被設定了,那麼x要嘛被設置,要嘛n ⊕ x被設置,因為1 ⊕ 1 = 0。這表示n ⊕ x沒有被設置,所以現在我們可以得出結論,對於n中的每個設定位,排列的數量為2^(設定位的數量)。
#include <bits/stdc++.h> using namespace std; int main (){ int n = 3, no_of_setbits = 0; // initialising n with value and taking count of set bits as 0 while (n != 0){ no_of_setbits = no_of_setbits + (n % 2); // checking if num contains set bit. n = n / 2; } int result = 1 << no_of_setbits; // calculating no. of possible solution with 2^setbits cout << " Number of possible solutions : " << result; return 0; }
Number of possible solutions : 4
The time complexity of this approach is O(n), as we are applying Brute force here. We can apply more efficient methods to improve the efficiency of the program.
In this article, we solve a problem to findve a problem to findve a number of solution −
n = x n ⊕ x. We also learned the C program for this problem and the complete approach by which we solved this problem. We can write the same program in other languages such as C, such as C, same program java, python, and other languages. Hope you find this article helpful.
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