将 N 个男孩和 M 个女孩放置在不同的排中，使得每排中放置的人数最大化
原文:https://www . geesforgeks . org/place-n-boys-m-girls-in-different-row-使得每一行的人数最大化/
给定两个整数 N 和 M 代表男孩和女孩的数量，任务是将它们排列在相同大小的不同行的数量中，使得每行包含尽可能多的学生，并且每行应该包含男孩或女孩。 注意:没有一行可以同时包含男生和女生。
例:
输入: N = 4，M = 2 输出: 2 说明:** 以下排列顺序满足给定条件: 1 st 行:B1，B2 2 nd 行:B3，B4 3 rd 行:G1，G2 清楚，每行
输入: N = 6，M = 6 输出: 6 说明: 以下排列顺序满足给定条件: 1 st 行:B1、B2、B3、B4、B5、B6 2 nd 行:G1、G2、G3、G4、G5、G6
方法:按照以下步骤解决问题
- 由于每行可以包含男孩或女孩，并且所有行的大小必须相同，因此最理想的排列方式是在每行中放置 (N，M) 数量的元素的大公约数。
- 因此，打印 GCD(N，M) 作为所需答案。
下面是上述方法的实现:
C++14
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate
// GCD of two numbers
int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to count maximum persons
// that can be placed in a row
int maximumRowValue(int n, int m) { return gcd(n, m); }
// Driver Code
int main()
{
// Input
int N = 4;
int M = 2;
// Function to count maximum
// persons that can be placed in a row
cout << maximumRowValue(N, M);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to calculate
// GCD of two numbers
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to count maximum persons
// that can be placed in a row
static int maximumRowValue(int n, int m) {
return gcd(n, m);
}
// Driver Code
public static void main(String args[])
{
// Input
int N = 4;
int M = 2;
// Function to count maximum
// persons that can be placed in a row
System.out.print(maximumRowValue(N, M));
}
}
// This code is contributed by SURENDRA_GANGWAR.
Python 3
# Python3 Program to implement
# the above approach
# Function to calculate
# GCD of two numbers
def gcd(a, b):
if (b == 0):
return a
return gcd(b, a % b)
# Function to count maximum persons
# that can be placed in a row
def maximumRowValue(n, m):
return gcd(n, m)
# Driver Code
if __name__ == '__main__':
# Input
N = 4
M = 2
# Function to count maximum
# persons that can be placed in a row
print (maximumRowValue(N, M))
# This code is contributed by mohit kumar 29.
C
// C# Program to implement
// the above approach
using System;
class GFG
{
// Function to calculate
// GCD of two numbers
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to count maximum persons
// that can be placed in a row
static int maximumRowValue(int n, int m) { return gcd(n, m); }
// Driver Code
public static void Main(String[] args)
{
// Input
int N = 4;
int M = 2;
// Function to count maximum
// persons that can be placed in a row
Console.WriteLine(maximumRowValue(N, M));
}
}
// This code is contributed by code_hunt.
java 描述语言
<script>
// javascript Program to implement
// the above approach
// Function to calculate
// GCD of two numbers
function gcd(a , b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to count maximum persons
// that can be placed in a row
function maximumRowValue(n , m) {
return gcd(n, m);
}
// Driver Code
// Input
var N = 4;
var M = 2;
// Function to count maximum
// persons that can be placed in a row
document.write(maximumRowValue(N, M));
// This code is contributed by aashish1995
</script>
Output:
2
时间复杂度: O(log N) 辅助 空间: O(1)
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