给定一个整数N ，任务是找到{1, 2, 5}值硬币的最小数量，以便可以形成范围[1, N]内所有可能值的变化，并且不可能获得值否。
例子：
Input: N = 8
Output:
Count of 5 values coins: 0
Count of 2 rupees coins: 3
Count of 1 rupees coins: 2
Explanation:
Coins required for 1 cent = 1
Coins required for 2 cent = 1
Coins required for 3 cent = 2
Coins required for 4 cent = 2
Coins required for 5 cent = 3
Coins required for 6 cent = 3
Coins required for 7 cent = 4
Coins required for 8 cent = 5
Input: N = 17
Output:
Count of 5 rupees coins: 2
Count of 2 rupees coins: 3
Count of 1 rupees coins: 1
方法：该问题可以使用贪心技术解决。该想法基于以下观察：
Any number, X from the range [1, N] can be represented as
X = 5 * (Integer) + Y * (Integer)
Y is one of the values from the range [0, 4]
请按照以下步骤解决问题：
- 初始化三个变量，说F，T，和O，来存储的5,2和1 -valued硬币的计数。
- 使用F = (N – 4)/5计算5 个价值硬币的数量。
- 如果 (N – 5 * F) 是偶数，那么一个有价硬币的数量可以计算为O = 1 。
- 否则，一枚有价硬币的数量可以计算为O = 2 。
- 计算两枚有价硬币的数量可以计算为T = (N – 5 * F – O) / 2。
- 最后，打印F、T和O 的值。
以下是上述方法的实现：，
C++
#include
using namespace std;
// Function to find minimum count of {1, 2, 5}
// valued coins required to make a change of
// all values in the range [1, N]
void find(int N)
{
int T, F, O;
// Number of 5 valueds coins required
F = int((N - 4) / 5);
// Number of 1 valued coins required
if (((N - 5 * F) % 2) == 0)
{
O = 2;
}
else
{
O = 1 ;
}
// Number of 2 valued coins required
T = floor((N - 5 * F - O)/2);
cout<< "Count of 5 valueds coins: " << F << endl;
cout<< "Count of 2 valueds coins: " << T<< endl;
cout<< "Count of 1 valueds coins: " << O << endl;
}
// Driver Code
int main()
{
int N = 8;
find(N);
return 0;
}
// This code is contributed by Jana_sayantan. Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to find minimum count of {1, 2, 5}
// valued coins required to make a change of
// all values in the range [1, N]
static void find(int N)
{
int T, F, O;
// Number of 5 valueds coins required
F = (int)((N - 4) / 5);
// Number of 1 valued coins required
if (((N - 5 * F) % 2) == 0)
{
O = 2;
}
else
{
O = 1 ;
}
// Number of 2 valued coins required
T = (int)Math.floor((N - 5 * F - O)/2);
System.out.println("Count of 5 valueds coins: " + F);
System.out.println("Count of 2 valueds coins: " + T);
System.out.println("Count of 1 valueds coins: " + O);
}
// Driver Code
public static void main(String args[])
{
int N = 8;
find(N);
}
}
// This code is contributed by splevel62.Python3
# Python Program for the above approach
# Function to find minimum count of {1, 2, 5}
# valued coins required to make a change of
# all values in the range [1, N]
def find(N):
# Number of 5 valueds coins required
F = int((N - 4) / 5)
# Number of 1 valued coins required
if ((N - 5 * F) % 2) == 0:
O = 2
else:
O = 1
# Number of 2 valued coins required
T = (N - 5 * F - O)//2
print("Count of 5 valueds coins: ", F)
print("Count of 2 valueds coins: ", T)
print("Count of 1 valueds coins: ", O)
if __name__ == '__main__':
N = 8
find(N)C#
// C# program to implement
// the above approach
using System;
public class GFG
{
// Function to find minimum count of {1, 2, 5}
// valued coins required to make a change of
// all values in the range [1, N]
static void find(int N)
{
int T, F, O;
// Number of 5 valueds coins required
F = (int)((N - 4) / 5);
// Number of 1 valued coins required
if (((N - 5 * F) % 2) == 0)
{
O = 2;
}
else
{
O = 1 ;
}
// Number of 2 valued coins required
T = (int)Math.Floor((double)(N - 5 * F - O)/2);
Console.WriteLine("Count of 5 valueds coins: " + F);
Console.WriteLine("Count of 2 valueds coins: " + T);
Console.WriteLine("Count of 1 valueds coins: " + O);
}
// Driver Code
public static void Main(String []args)
{
int N = 8;
find(N);
}
}
// This code is contributed by 29AjayKumarJavascript
输出：
Count of 5 valueds coins: 0
Count of 2 valueds coins: 3
Count of 1 valueds coins: 2时间复杂度： O(1)
辅助空间： O(1)