按照给定的条件
是否可以从 1 和 0 分别达到 N 和 M
原文:https://www . geesforgeks . org/根据给定条件分别从 1 和 0 到达 n 和 m 是可能的吗/
给定两个整数 N 和 M ，任务是检查是否可以通过多次执行这两个操作分别从 X = 1 和 Y = 0 获得这些值:
- 当且仅当 x>0 时，将 X 和 Y 增加 1。
- 当且仅当 y>0 时，将 Y 增加 2。
示例:
输入: N = 3，M = 4 输出:是 解释: 最初 X = 1，Y = 0 运算 1: X = 2，Y = 1 运算 1: X = 3，Y = 2 运算 2: X = 3，Y = 4，由此得到最终值，所以答案是肯定的。
输入: N = 5，M = 2 输出:否 说明: 从 X = 1 和 Y = 0 获得 X = 5 和 Y = 2 是不可能的。
方法:上述问题可以通过以下观察来解决:
- 如果 N 小于 2M 不等于零，则无法得到最终值，因此答案为否。
- 否则，从 M 中减去 N，如果 M？0 和 M 可被 2 整除那么答案是是。
- 在所有其他情况下，答案是否。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that find given x and y
// is possible or not
bool is_possible(int x, int y)
{
// Check if x is less than 2 and
// y is not equal to 0
if (x < 2 && y != 0)
return false;
// Perform subtraction
y = y - x + 1;
// Check if y is divisible by 2
// and greater than equal to 0
if (y % 2 == 0 && y >= 0)
return true;
else
return false;
}
// Driver Code
int main()
{
// Given X and Y
int x = 5, y = 2;
// Function Call
if (is_possible(x, y))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
class GFG{
// Function that find given x and y
// is possible or not
static boolean is_possible(int x, int y)
{
// Check if x is less than 2 and
// y is not equal to 0
if (x < 2 && y != 0)
return false;
// Perform subtraction
y = y - x + 1;
// Check if y is divisible by 2
// and greater than equal to 0
if (y % 2 == 0 && y >= 0)
return true;
else
return false;
}
// Driver Code
public static void main(String[] args)
{
// Given X and Y
int x = 5, y = 2;
// Function Call
if (is_possible(x, y))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by rock_cool
Python 3
# Python3 program for the above approach
# Function that find given x and y
# is possible or not
def is_possible(x, y):
# Check if x is less than 2 and
# y is not equal to 0
if (x < 2 and y != 0):
return false
# Perform subtraction
y = y - x + 1
# Check if y is divisible by 2
# and greater than equal to 0
if (y % 2 == 0 and y >= 0):
return True
else:
return False
# Driver Code
if __name__ == '__main__':
# Given X and Y
x = 5
y = 2
# Function Call
if (is_possible(x, y)):
print("Yes")
else:
print("No")
# This code is contributed by Mohit Kumar
C
// C# program for the above approach
using System;
class GFG{
// Function that find given x and y
// is possible or not
static bool is_possible(int x, int y)
{
// Check if x is less than 2 and
// y is not equal to 0
if (x < 2 && y != 0)
return false;
// Perform subtraction
y = y - x + 1;
// Check if y is divisible by 2
// and greater than equal to 0
if (y % 2 == 0 && y >= 0)
return true;
else
return false;
}
// Driver Code
public static void Main(string[] args)
{
// Given X and Y
int x = 5, y = 2;
// Function Call
if (is_possible(x, y))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Ritik Bansal
java 描述语言
<script>
// Javascript program for the above approach
// Function that find given x and y
// is possible or not
function is_possible(x, y)
{
// Check if x is less than 2 and
// y is not equal to 0
if (x < 2 && y != 0)
return false;
// Perform subtraction
y = y - x + 1;
// Check if y is divisible by 2
// and greater than equal to 0
if (y % 2 == 0 && y >= 0)
return true;
else
return false;
}
// Driver code
// Given X and Y
let x = 5, y = 2;
// Function Call
if (is_possible(x, y))
document.write("Yes");
else
document.write("No");
// This code is contributed by divyesh072019
</script>
Output:
No
时间复杂度: O(1) 辅助空间: O(1)
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