检查是否可以构造一个大小为 N 的数组，其总和为 S，XOR 值为 X
给定三个数字N、S和X ，任务是找出是否有可能构造一个长度为N的序列A ，其中每个A[i] >= 0表示1<=i<=N以及所有数字的总和一个序列等于S ，序列的按位异或等于X 。
例子：
Input: N = 3, S = 10, X = 4
Output: Yes
Explanation: One of the sequence possible is {4, 3, 3} where sum equals 10 and XOR equals 4
Input: N = 1, S = 5, X = 3
Output: No
方法：让我们考虑以下测试用例。
Case-1：当N等于1时，很容易看出，当(S等于X)时，只返回“是”，否则返回“否”。
Case-2：当N大于等于3时，使用公式(a + b) = (a xor b) + 2(a and b)这里可以看出(a + b) = S and (a xor b) = X所以方程变为S = X + 2(ab)。因此， (SX)应该是偶数，因为在右侧我们有2(ab)。所以，可以说是 S 是奇数，那么 X 是奇数，如果 S 是偶数，那么 X 是偶数，那么只有 (SX) 也是偶数，这可以通过(S%2 == X%2)也S >= X否则 AB 变为负数，这是不可能的。
案例 3：对于案例N等于3，它类似于A + B + C = S和A^B^C = X。使用属性A^A = 0和0^A = A => X + ( S – X)/2 + (S – X)/2 = X + (SX) => X + ( S – X)/2 + (S – X)/2 = S也是这样： X ^( ( S – X)/2 ^ (SX)/2 ) = X ^ 0 = X。因此，证明对于N == 3总会有这样的序列，我们可以返回“是”。
案例 4：当N == 2且(S%2 == X%2)且S >= X时，假设A + B == S且(A^B) == X则(A 和 B) == (SX)/2从上面讨论的等式。令C = AB仔细观察可以注意到，只有当 A 和 B 位在该位置为“1”时，C 的位才为“1”，否则为“1”。并且 X 是 A 的异或，B 只有在第 i个位置有不同的位时 A 有 '0' 而 B 有 '1' 或正好相反：所以看看这个序列，将每个位分配到变量A和B，C = ( S – X)/2。从 C 分配 A 和 B -> A = C, B = C
现在将X添加到A或B以将所有 1 分配给A并将所有 0 分配给B所以当我们对两个数字进行异或时，添加到 A 中的“1”位将与我们添加到 B 中的“0”相反.有趣的部分是当 C 的设置位与 X 的一些设置位重合时，它不会给出 X 的所需 xor，现在， A = C + X，B = C。现在A+B = ( C + X) + C = S并且当 XOR AB 等于 X 时可以确定存在这样的对，当A + B == S和(A^B) == X 时；
请按照以下步骤解决问题：
- 如果S大于等于X，并且S%2等于X%2则执行以下步骤，否则返回No。
- 如果n大于等于3，则返回Yes。
- 如果n等于1，如果S等于X，则返回Yes ，否则返回No。
- 如果n等于2，则将变量C初始化为(SX)/2 ，并将变量A和B设置为C ，并将值X添加到变量A ，如果A^B等于X，则打印Yes ，否则打印No。
下面是上述方法的实现。
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find if any sequence is
// possible or not.
string findIfPossible(int N, int S, int X)
{
if (S >= X and S % 2 == X % 2) {
// Since, S is greater than equal to
// X, and either both are odd or even
// There always exists a sequence
if (N >= 3) {
return "Yes";
}
if (N == 1) {
// Only one case possible is
// S == X or NOT;
if (S == X) {
return "Yes";
}
else {
return "No";
}
}
// Considering the above conditions true,
// check if XOR of S^(S-X) is X or not
if (N == 2) {
int C = (S - X) / 2;
int A = C;
int B = C;
A = A + X;
if (((A ^ B) == X)) {
return "Yes";
}
else {
return "No";
}
}
}
else {
return "No";
}
}
// Driver Code
int main()
{
int N = 3, S = 10, X = 4;
cout << findIfPossible(N, S, X);
return 0;
} C
// C program for the above approach
#include
#include
#include
// Function to find if any sequence is
// possible or not.
char* findIfPossible(int N, int S, int X)
{
if (S >= X && S % 2 == X % 2) {
// Since, S is greater than equal to
// X, and either both are odd or even
// There always exists a sequence
if (N >= 3) {
return "Yes";
}
if (N == 1) {
// Only one case possible is
// S == X or NOT;
if (S == X) {
return "Yes";
}
else {
return "No";
}
}
// Considering the above conditions true,
// check if XOR of S^(S-X) is X or not
if (N == 2) {
int C = (S - X) / 2;
int A = C;
int B = C;
A = A + X;
if (((A ^ B) == X)) {
return "Yes";
}
else {
return "No";
}
}
}
else {
return "No";
}
}
// Driver Code
int main()
{
int N = 3, S = 10, X = 4;
printf("%s\n", findIfPossible(N, S, X));
return 0;
}
// This code is contributed by phalasi. Java
// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find if any sequence is
// possible or not.
static void findIfPossible(int N, int S, int X)
{
if ((S >= X) && (S % 2 == X % 2)) {
// Since, S is greater than equal to
// X, and either both are odd or even
// There always exists a sequence
if (N >= 3) {
System.out.println("Yes");
}
if (N == 1) {
// Only one case possible is
// S == X or NOT;
if (S == X) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
// Considering the above conditions true,
// check if XOR of S^(S-X) is X or not
if (N == 2) {
int C = (S - X) / 2;
int A = C;
int B = C;
A = A + X;
if (((A ^ B) == X)) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
}
else {
System.out.println("No");
}
}
// Driver code
public static void main(String args[])
{
int N = 3, S = 10, X = 4;
findIfPossible(N, S, X);
}
}
// This code is contributed by code_hunt.Python3
# Python program for the above approach
# Function to find if any sequence is
# possible or not.
def findIfPossible(N, S, X):
if (S >= X and S % 2 == X % 2):
# Since, S is greater than equal to
# X, and either both are odd or even
# There always exists a sequence
if (N >= 3):
return "Yes"
if (N == 1):
# Only one case possible is
# S == X or NOT
if (S == X):
return "Yes"
else:
return "No"
# Considering the above conditions true,
# check if XOR of S^(S-X) is X or not
if (N == 2):
C = (S - X) // 2
A = C
B = C
A = A + X
if (((A ^ B) == X)):
return "Yes"
else:
return "No"
else:
return "No"
# Driver Code
N = 3
S = 10
X = 4
print(findIfPossible(N, S, X))
# This code is contributed by shivanisinghss2110C#
// C# program for the above approach
using System;
public class GFG {
// Function to find if any sequence is
// possible or not.
static void findIfPossible(int N, int S, int X)
{
if ((S >= X) && (S % 2 == X % 2)) {
// Since, S is greater than equal to
// X, and either both are odd or even
// There always exists a sequence
if (N >= 3) {
Console.WriteLine("Yes");
}
if (N == 1) {
// Only one case possible is
// S == X or NOT;
if (S == X) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
// Considering the above conditions true,
// check if XOR of S^(S-X) is X or not
if (N == 2) {
int C = (S - X) / 2;
int A = C;
int B = C;
A = A + X;
if (((A ^ B) == X)) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
}
else {
Console.WriteLine("No");
}
}
// Driver code
public static void Main(String[] args)
{
int N = 3, S = 10, X = 4;
findIfPossible(N, S, X);
}
}
// This code is contributed by Princi SinghJavascript
Yes时间复杂度： O(1)
辅助空间： O(1)