给定两个整数A和B (代表两种不同类型的对象的数量)，另一个整数N (代表架子的数量)，任务是按照以下规则将所有对象放置在给定的N个架子中：
- 任何架子都不能同时包含Type-A和Type-B对象。
- 架子最多只能容纳K个A型对象或L个B型对象。
如果可以将所有物品放置在N个货架上，则打印“是” 。否则，打印“否” 。
例子：
Input: A = 3, B = 3, N = 3, K = 4, M = 2
Output: YES
Explanation:
3 Type-A items can be placed on 1 shelf, as maximum limit is 4.
3 Type-B items can be placed on 2 shelves, as maximum limit is 2.
Since the required number of shelves does not exceed N, so allocation is possible.
Input: A = 6, B = 7, N = 3, K = 4, L = 5
Output: NO
Explaination:
6 Type-A items require 2 shelves, as maximum limit is 4.
7 Type-B items require 2 shelves, as maximum limit is 5.
Since the required number of shelves exceeds N, so allocation is not possible.
方法：
为了解决该问题，我们需要计算放置所有物体所需的最小货架数，并检查其是否超过N。请按照以下步骤操作：
- 计算放置Type-A项目所需的最小项目数，例如needa 。由于K个A型物品最多只能放在一个架子上，因此会出现以下两种情况：
- 如果A可被K整除，则所有Type-A物品都可以放在A / K货架上。
- 否则，必须将A％K的物品放在一个架子上，其余的放在A / K架子上，因此这种情况下需要A / K + 1个架子。
- 同样，计算放置B型物品所需的最小货架数，例如Needb 。
- 如果Needa + Needb超过N ，则无法分配。否则，这是可能的。
下面是上述方法的实现。
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to return if allocation
// is possible or not
bool isPossible(int A, int B, int N,
int K, int L)
{
// Stores the shelves needed
// for items of type-A and type-B
int needa, needb;
// Find number of shelves
// needed for items of type-A
if (A % K == 0)
// Fill A / K shelves fully
// by the items of type-A
needa = A / K;
// Otherwise
else
// Fill A / L shelves fully
// and add remaining to an
// extra shelf
needa = A / K + 1;
// Find number of shelves
// needed for items of type-B
if (B % L == 0)
// Fill B / L shelves fully
// by the items of type-B
needb = B / L;
else
// Fill B / L shelves fully
// and add remaining to an
// an extra shelf
needb = B / L + 1;
// Total shelves needed
int total = needa + needb;
// If required shelves exceed N
if (total > N)
return false;
else
return true;
}
// Driver Program
int main()
{
int A = 3, B = 3, N = 3;
int K = 4, M = 2;
if (isPossible(A, B, N, K, M))
cout << "YES" << endl;
else
cout << "NO" << endl;
return 0;
} Java
// Java implementation of the above approach
class GFG{
// Function to return if allocation
// is possible or not
static boolean isPossible(int A, int B,
int N, int K,
int L)
{
// Stores the shelves needed
// for items of type-A and type-B
int needa, needb;
// Find number of shelves
// needed for items of type-A
if (A % K == 0)
// Fill A / K shelves fully
// by the items of type-A
needa = A / K;
// Otherwise
else
// Fill A / L shelves fully
// and add remaining to an
// extra shelf
needa = A / K + 1;
// Find number of shelves
// needed for items of type-B
if (B % L == 0)
// Fill B / L shelves fully
// by the items of type-B
needb = B / L;
else
// Fill B / L shelves fully
// and add remaining to an
// an extra shelf
needb = B / L + 1;
// Total shelves needed
int total = needa + needb;
// If required shelves exceed N
if (total > N)
return false;
else
return true;
}
// Driver code
public static void main(String[] args)
{
int A = 3, B = 3, N = 3;
int K = 4, M = 2;
if (isPossible(A, B, N, K, M))
System.out.print("YES" + "\n");
else
System.out.print("NO" + "\n");
}
}
// This code is contributed by amal kumar choubeyPython3
# Python3 implementation of the
# above approach
# Function to return if allocation
# is possible or not
def isPossible(A, B, N, K, L):
# Stores the shelves needed
# for items of type-A and type-B
needa = 0
needb = 0
# Find number of shelves
# needed for items of type-A
if (A % K == 0):
# Fill A / K shelves fully
# by the items of type-A
needa = A // K;
# Otherwise
else:
# Fill A / L shelves fully
# and add remaining to an
# extra shelf
needa = A // K + 1
# Find number of shelves
# needed for items of type-B
if (B % L == 0):
# Fill B / L shelves fully
# by the items of type-B
needb = B // L
else:
# Fill B / L shelves fully
# and add remaining to an
# an extra shelf
needb = B // L + 1
# Total shelves needed
total = needa + needb
# If required shelves exceed N
if (total > N):
return False
else:
return True
# Driver Code
if __name__=='__main__':
A, B, N = 3, 3, 3
K, M = 4, 2
if (isPossible(A, B, N, K, M)):
print('YES')
else:
print('NO')
# This code is contributed by rutvik_56C#
// C# implementation of the above approach
using System;
class GFG{
// Function to return if allocation
// is possible or not
static bool isPossible(int A, int B,
int N, int K,
int L)
{
// Stores the shelves needed
// for items of type-A and type-B
int needa, needb;
// Find number of shelves
// needed for items of type-A
if (A % K == 0)
// Fill A / K shelves fully
// by the items of type-A
needa = A / K;
// Otherwise
else
// Fill A / L shelves fully
// and add remaining to an
// extra shelf
needa = A / K + 1;
// Find number of shelves
// needed for items of type-B
if (B % L == 0)
// Fill B / L shelves fully
// by the items of type-B
needb = B / L;
else
// Fill B / L shelves fully
// and add remaining to an
// an extra shelf
needb = B / L + 1;
// Total shelves needed
int total = needa + needb;
// If required shelves exceed N
if (total > N)
return false;
else
return true;
}
// Driver code
public static void Main(String[] args)
{
int A = 3, B = 3, N = 3;
int K = 4, M = 2;
if (isPossible(A, B, N, K, M))
Console.Write("YES" + "\n");
else
Console.Write("NO" + "\n");
}
}
// This code is contributed by Rohit_ranjan输出：
YES
时间复杂度： O(1)
辅助空间： O(1)