给定A号白球和B号黑球。您需要有X个白球和Y个黑球（A <= X，B <= Y）才能通过执行一些（零个或多个）操作来赢得比赛。
在一个操作中：在任何时候，如果您有 p 个白球和 q 个黑球，那么在那一刻您可以购买 q 个白球或 p 个黑球。
找出最后是否有可能有 X 个白球和 Y 个黑球。
例子：
Input: A = 1, B = 1, X = 3, Y = 8
Output: POSSIBLE
Explanation:
The steps are, (1, 1)->(1, 2)->(3, 2)->(3, 5)->(3, 8)
Input: A = 3, Y = 2, X = 4, Y = 6
Output: NOT POSSIBLE方法：
我们必须使用 gcd 的属性来解决这个问题。让我们看看如何。
- 最初，我们有 A 白球和 B 黑球。我们必须得到其余的 XA 红球和 YB 黑球。
- 下面是我们将使用的两个数字的 gcd 属性，
gcd(x, y) = gcd(x + y, y)
gcd(x, y) = gcd(x, y + x)- 此属性与问题中提到的操作相同。所以从这里，我们得到如果最终状态的 gcd 与初始状态的 gcd 相同，那么总是有可能达到目标，否则不会。
下面是上述方法的实现：
C++
// C++ program to Find is it possible
// to have X white and Y black
// balls at the end.
#include
using namespace std;
// Recursive function to return
// gcd of a and b
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function returns if it's
// possible to have X white
// and Y black balls or not.
void IsPossible(int a, int b,
int x, int y)
{
// Finding gcd of (x, y)
// and (a, b)
int final = gcd(x, y);
int initial = gcd(a, b);
// If gcd is same, it's always
// possible to reach (x, y)
if (initial == final)
{
cout << "POSSIBLE\n";
}
else
{
// Here it's never possible
// if gcd is not same
cout << "NOT POSSIBLE\n";
}
}
// Driver Code
int main()
{
int A = 1, B = 2, X = 4, Y = 11;
IsPossible(A, B, X, Y);
A = 2, B = 2, X = 3, Y = 6;
IsPossible(A, B, X, Y);
return 0;
} Java
// Java program to Find is it possible
// to have X white and Y black
// balls at the end.
import java.io.*;
class GFG{
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function returns if it's
// possible to have X white
// and Y black balls or not.
static void IsPossible(int a, int b,
int x, int y)
{
// Finding gcd of (x, y)
// and (a, b)
int g = gcd(x, y);
int initial = gcd(a, b);
// If gcd is same, it's always
// possible to reach (x, y)
if (initial == g)
{
System.out.print("POSSIBLE\n");
}
else
{
// Here it's never possible
// if gcd is not same
System.out.print("NOT POSSIBLE\n");
}
}
// Driver code
public static void main(String args[])
{
int A = 1, B = 2, X = 4, Y = 11;
IsPossible(A, B, X, Y);
A = 2; B = 2; X = 3; Y = 6;
IsPossible(A, B, X, Y);
}
}
// This code is contributed by shivanisinghss2110Python3
# Python3 program to find is it possible
# to have X white and Y black
# balls at the end.
# Recursive function to return
# gcd of a and b
def gcd(a, b) :
if (b == 0) :
return a;
return gcd(b, a % b);
# Function returns if it's
# possible to have X white
# and Y black balls or not.
def IsPossible(a, b, x, y) :
# Finding gcd of (x, y)
# and (a, b)
final = gcd(x, y);
initial = gcd(a, b);
# If gcd is same, it's always
# possible to reach (x, y)
if (initial == final) :
print("POSSIBLE");
else :
# Here it's never possible
# if gcd is not same
print("NOT POSSIBLE");
# Driver Code
if __name__ == "__main__" :
A = 1; B = 2; X = 4; Y = 11;
IsPossible(A, B, X, Y);
A = 2; B = 2; X = 3; Y = 6;
IsPossible(A, B, X, Y);
# This code is contributed by AnkitRai01C#
// C# program to Find is it possible
// to have X white and Y black
// balls at the end.
using System;
using System.Linq;
class GFG {
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function returns if it's
// possible to have X white
// and Y black balls or not.
static void IsPossible(int a, int b,
int x, int y)
{
// Finding gcd of (x, y)
// and (a, b)
int g = gcd(x, y);
int initial = gcd(a, b);
// If gcd is same, it's always
// possible to reach (x, y)
if (initial == g)
{
Console.Write("POSSIBLE\n");
}
else
{
// Here it's never possible
// if gcd is not same
Console.Write("NOT POSSIBLE\n");
}
}
// Driver code
static public void Main()
{
int A = 1, B = 2;
int X = 4, Y = 11;
IsPossible(A, B, X, Y);
A = 2; B = 2;
X = 3; Y = 6;
IsPossible(A, B, X, Y);
}
}
// This code is contributed by shivanisinghss2110Javascript
输出：
POSSIBLE
NOT POSSIBLE如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。