从它们的和、差、积、除和余数中找到 X 和 Y
给定5 个整数的arr[] ，表示 X+Y、X−Y、X*Y、X%Y 和 ⌊X/Y⌋ 对于两个非零整数 X 和 Y 的排序顺序，任务是找到X 和 Y 的值。
注意：如果不存在解决方案，则返回两个 0
例子：
Input: -1, 0, 4, 9, 20
Output: X = 4, Y = 5
Explanation: If we consider, X + Y = 9, X – Y = -1.
Then X and Y are 4 and 5, which satisfies the condition
X * Y = 20, X % Y = 4, Floor(X / Y) = 0.
Input: -3, -1, 0, 2, 2
Output: X = -2, Y = -1
方法：可以根据以下数学观察解决问题
If we have X+Y and X-Y we can find X and Y.
X = [(X+Y)+(X-Y)]/2
Y = X+Y-X
为了实现上述想法，使用回溯尝试所有可能的X+Y和XY值对。请按照以下步骤解决问题：
- 将数组元素视为 A、B、C、D、E
- 尝试所有可能的 X+Y 和 XY 值对。
- 根据上述观察，从这两个值中找到 X 和 Y。
- 如果 X 和 Y 的这两个值满足其他值，则返回。
- 否则，请尝试另一对。
下面是上述方法的实现：
C++
// C++ code to implement the approach
#include
#define ll long long
using namespace std;
// Function to check if X and Y
// are valid or not
pair isValid(ll A, ll B, ll C, ll D, ll E)
{
// a represents 2*a for now
ll a = A + B;
// 2a/2 = a that must be integer
if (ceil(float(a / 2)) != floor(float(a / 2))) {
return make_pair(0, 0);
}
else {
// Find value of a
a = a / 2;
// Find value of b
ll b = A - a;
// Edge Cases
if (a == 0 || b == 0)
return make_pair(0, 0);
else if ((a + b) > pow(10, 3)
|| (a - b) < pow(-10, 3))
return make_pair(0, 0);
// 1st Condition, C = a*b
else if ((a * b == C) && (a / b == D)
&& (a % b == E)
|| (a * b == C) && (a / b == E)
&& (a % b == D))
return make_pair(a, b);
// 2nd Condition, D = a*b
else if ((a * b == D) && (a / b == C)
&& (a % b == E)
|| (a * b == D) && (a / b == E)
&& (a % b == C))
return make_pair(a, b);
// 3rd Condition, E = a*b
else if ((a * b == E) && (a / b == C)
&& (a % b == D)
|| (a * b == E) && (a / b == D)
&& (a % b == C))
return make_pair(a, b);
// Pairs are not valid then return 0
else
return make_pair(0, 0);
}
}
// Function to find two integers X and Y
void findNum(ll* arr)
{
pair p;
bool flag = 0;
for (int i = 0; i <= 4; i++) {
// Swaping for every
// X + Y combination
swap(arr[0], arr[i]);
for (int j = 1; j <= 4; j++) {
// Swaping for every
// X - Y combination
swap(arr[1], arr[j]);
// Checking for valid X and Y
p = isValid(arr[0], arr[1],
arr[2], arr[3],
arr[4]);
// If both are not -1 then
// we found X and Y
if ((p.first != 0)
&& (p.second != 0)) {
// Set Flag = true
flag = 1;
// Print the values in order
// i.e., X and Y
cout << p.first << " "
<< p.second << endl;
}
// Backtracking
swap(arr[1], arr[j]);
// X and Y are found
if (flag)
break;
}
// Backtracking
swap(arr[0], arr[i]);
// X and Y are found
if (flag)
break;
}
// If flag is 0 then X and Y
// can't be possible
if (!flag)
cout << 0 << " " << 0 << endl;
}
// Driver Code
int main()
{
int N = 5;
ll arr[N] = { -1, 0, 4, 9, 20 };
// Function call
findNum(arr);
return 0;
} Java
// Java code to implement the approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to check if X and Y
// are valid or not
public static long[] isValid(long A, long B, long C,
long D, long E)
{
// a represents 2*a for now
long a = A + B;
// 2a/2 = a that must be integer
if (Math.ceil((a / 2.0)) != Math.floor((a / 2.0))) {
long ans[] = { 0, 0 };
return ans;
}
else {
// Find value of a
a = a / 2;
// Find value of b
long b = A - a;
long res[] = { 0, 0 };
long res1[] = { a, b };
// Edge Cases
if (a == 0 || b == 0)
return res;
else if ((a + b) > Math.pow(10, 3)
|| (a - b) < Math.pow(-10, 3))
return res;
// 1st Condition, C = a*b
else if ((a * b == C) && (a / b == D)
&& (a % b == E)
|| (a * b == C) && (a / b == E)
&& (a % b == D))
return res1;
// 2nd Condition, D = a*b
else if ((a * b == D) && (a / b == C)
&& (a % b == E)
|| (a * b == D) && (a / b == E)
&& (a % b == C))
return res1;
// 3rd Condition, E = a*b
else if ((a * b == E) && (a / b == C)
&& (a % b == D)
|| (a * b == E) && (a / b == D)
&& (a % b == C))
return res1;
// Pairs are not valid then return 0
else
return res;
}
}
// Function to find two integers X and Y
public static void findNum(long arr[])
{
long p[] = new long[2];
int flag = 0;
for (int i = 0; i <= 4; i++) {
// Swaping for every
// X + Y combination
long tmp = arr[0];
arr[0] = arr[i];
arr[i] = tmp;
for (int j = 1; j <= 4; j++) {
// Swaping for every
// X - Y combination
tmp = arr[1];
arr[1] = arr[j];
arr[j] = tmp;
// Checking for valid X and Y
p = isValid(arr[0], arr[1], arr[2], arr[3],
arr[4]);
// If both are not -1 then
// we found X and Y
if ((p[0] != 0) && (p[1] != 0)) {
// Set Flag = true
flag = 1;
// Print the values in order
// i.e., X and Y
System.out.println(p[0] + " " + p[1]);
}
// Backtracking
tmp = arr[1];
arr[1] = arr[j];
arr[j] = tmp;
// X and Y are found
if (flag != 0)
break;
}
// Backtracking
tmp = arr[0];
arr[0] = arr[i];
arr[i] = tmp;
// X and Y are found
if (flag != 0)
break;
}
// If flag is 0 then X and Y
// can't be possible
if (flag == 0)
System.out.println("0 0");
}
public static void main(String[] args)
{
int N = 5;
long arr[] = { -1, 0, 4, 9, 20 };
// Function call
findNum(arr);
}
}
// This code is contributed by Rohit PradhanC#
// C# code to implement the approach
using System;
public class GFG{
// Function to check if X and Y
// are valid or not
static long[] isValid(long A, long B, long C,
long D, long E)
{
// a represents 2*a for now
long a = A + B;
// 2a/2 = a that must be integer
if (Math.Ceiling((a / 2.0)) != Math.Floor((a / 2.0))) {
long[] ans = { 0, 0 };
return ans;
}
else {
// Find value of a
a = a / 2;
// Find value of b
long b = A - a;
long[] res = { 0, 0 };
long[] res1 = { a, b };
// Edge Cases
if (a == 0 || b == 0)
return res;
else if ((a + b) > Math.Pow(10, 3)
|| (a - b) < Math.Pow(-10, 3))
return res;
// 1st Condition, C = a*b
else if ((a * b == C) && (a / b == D)
&& (a % b == E)
|| (a * b == C) && (a / b == E)
&& (a % b == D))
return res1;
// 2nd Condition, D = a*b
else if ((a * b == D) && (a / b == C)
&& (a % b == E)
|| (a * b == D) && (a / b == E)
&& (a % b == C))
return res1;
// 3rd Condition, E = a*b
else if ((a * b == E) && (a / b == C)
&& (a % b == D)
|| (a * b == E) && (a / b == D)
&& (a % b == C))
return res1;
// Pairs are not valid then return 0
else
return res;
}
}
// Function to find two integers X and Y
static void findNum(long[] arr)
{
long[] p = new long[2];
int flag = 0;
for (int i = 0; i <= 4; i++) {
// Swaping for every
// X + Y combination
long tmp = arr[0];
arr[0] = arr[i];
arr[i] = tmp;
for (int j = 1; j <= 4; j++) {
// Swaping for every
// X - Y combination
tmp = arr[1];
arr[1] = arr[j];
arr[j] = tmp;
// Checking for valid X and Y
p = isValid(arr[0], arr[1], arr[2], arr[3],
arr[4]);
// If both are not -1 then
// we found X and Y
if ((p[0] != 0) && (p[1] != 0)) {
// Set Flag = true
flag = 1;
// Print the values in order
// i.e., X and Y
Console.WriteLine(p[0] + " " + p[1]);
}
// Backtracking
tmp = arr[1];
arr[1] = arr[j];
arr[j] = tmp;
// X and Y are found
if (flag != 0)
break;
}
// Backtracking
tmp = arr[0];
arr[0] = arr[i];
arr[i] = tmp;
// X and Y are found
if (flag != 0)
break;
}
// If flag is 0 then X and Y
// can't be possible
if (flag == 0)
Console.WriteLine("0 0");
}
static public void Main (){
long[] arr = { -1, 0, 4, 9, 20 };
// Function call
findNum(arr);
}
}
// This code is contributed by hrithikgarg03188.输出
4 5时间复杂度：O(1)
辅助空间：O(1)