给定无限数量的两种面额X和Y的硬币。还赠送了容量为N卢比的袋子，与硬币数量无关。任务是找到最小数量的袋子，以使每个袋子包含相同数量的卢比，并且所有袋子的总和至少为M。
例子 ：
Input : M = 27, N = 12, X = 2, Y = 5.
Output : 3
We put 2 coins of X, 1 coin of Y in each bag.
So we have 9 rupees in each bag and we need
at least 3 bags (Note that 27/9 = 3). There
is no way to obtain sum with lesser number
of bags.
Input : M = 45, N = 9, X = 4, Y = 5.
Output : 5任务是最小化袋子的数量，因此需要最大化袋子中的数量，以使所有袋子中的数量相同。假设我们采用x个硬币的“ p”个数量和Y个硬币的“ q”个数量，那么任务是使p * X + q * Y最大化。而且，p * X + q * Y <=N。
现在，要找到方程式左手侧的最大可能值，请将p从0更改为N / X，然后找到特定p的最大可能q。然后，在所有此类对中，取(p，q)对，其对为p * X + q * Y的最大值。
下面是上述想法的实现：
C++
// C++ program to find minimum number of bags such
// that each bag contains same amount and sum is at
// least M.
#include
using namespace std;
// Return minimum number of bags required.
int minBags(int M, int N, int X, int Y)
{
// Initialize maximum amount in a bag
int maxAmount = 0;
// Finding maximum possible q for the particular p.
for (int p = 0; p <= N/X; p++)
{
int q = (N - p * X) / Y;
maxAmount = max(maxAmount, p*X + q*Y);
}
// Calculating the minimum number of bags.
int result = M/maxAmount;
result += (M % maxAmount == 0? 0: 1);
return result;
}
// Driven Program
int main()
{
int M = 45, N = 9;
int X = 4, Y = 5;
cout << minBags(M, N, X, Y) << endl;
return 0;
} Java
// Java program to find minimum number
// of bags such that each bag contains
// same amount and sum is at least M
import java.io.*;
public class GFG {
// Return minimum number of bags required.
static int minBags(int M, int N,
int X, int Y)
{
// Initialize maximum amount in a bag
int maxAmount = 0;
// Finding maximum possible q
// for the particular p.
for (int p = 0; p <= N / X; p++)
{
int q = (N - p * X) / Y;
maxAmount = Math.max(maxAmount, p * X +
q * Y);
}
// Calculating the minimum number of bags.
int result = M / maxAmount;
result += (M % maxAmount == 0? 0: 1);
return result;
}
// Driver Code
static public void main (String[] args)
{
int M = 45, N = 9;
int X = 4, Y = 5;
System.out.println(minBags(M, N, X, Y));
}
}
// This code is contributed by vt_m.Python3
# Python 3 program to find minimum number
# of bags such that each bag contains same
# amount and sum is at least M.
# Return minimum number of bags required.
def minBags(M, N, X, Y):
# Initialize maximum amount in a bag
maxAmount = 0
# Finding maximum possible q for
# the particular p.
for p in range(0, int(N / X) + 1, 1):
q = int((N - p * X) / Y)
maxAmount = max(maxAmount, p * X + q * Y)
# Calculating the minimum number of bags.
result = int(M / maxAmount)
if(M % maxAmount == 0):
result += 0
else:
result += 1
return result
# Driver Code
if __name__ == '__main__':
M = 45
N = 9
X = 4
Y = 5
print(minBags(M, N, X, Y))
# This code is contributed by
# Surendra_GangwarC#
// C# program to find minimum number of
// bags such that each bag contains same
// amount and sum is at least M.
using System;
public class GFG
{
// Return minimum number of bags required.
static int minBags(int M, int N,
int X, int Y)
{
// Initialize maximum amount in a bag
int maxAmount = 0;
// Finding maximum possible q
// for the particular p.
for (int p = 0; p <= N / X; p++)
{
int q = (N - p * X) / Y;
maxAmount = Math.Max(maxAmount, p * X +
q * Y);
}
// Calculating the minimum number of bags.
int result = M / maxAmount;
result += (M % maxAmount == 0? 0: 1);
return result;
}
// Driver Code
static public void Main ()
{
int M = 45, N = 9;
int X = 4, Y = 5;
Console.WriteLine(minBags(M, N, X, Y));
}
}
// This code is contributed by vt_m.PHP
Javascript
输出 ：
5