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📜  总数为S的最小数字(小于或等于N)

📅  最后修改于: 2021-05-04 19:48:05             🧑  作者: Mango

给定N个数字(1、2、3,…. N)和一个数字S。任务是打印总和为S的最小数量的数字。

例子

方法:尽可能地选择N,然后如果剩余少于N的数字,我们将选择该数字加起来得到S,因此,数字的总数为(S / N)+1(如果S %N> 0)

下面是上述方法的实现。

C++
// C++ program to find the minimum numbers
// required to get to S
#include 
using namespace std;
  
// Function to find the minimum
// numbers required to get to S
int minimumNumbers(int n, int s)
{
    if (s % n)
        return s / n + 1;
    else
        return s / n;
}
  
// Driver Code
int main()
{
    int n = 5;
    int s = 11;
    cout << minimumNumbers(n, s);
    return 0;
}


Java
// Java program to find the minimum numbers
// required to get to S
import java.io.*;
  
class GFG {
      
  
// Function to find the minimum
// numbers required to get to S
static int minimumNumbers(int n, int s)
{
    if ((s % n)>0)
        return s / n + 1;
    else
        return s / n;
}
  
// Driver Code
  
    public static void main (String[] args) {
        int n = 5;
    int s = 11;
    System.out.println(minimumNumbers(n, s));
    }
}
// This code is contributed by shs..


Python 3
# Python 3 program to find the 
# minimum numbers required to get to S
  
# Function to find the minimum
# numbers required to get to S
def minimumNumbers(n, s):
  
    if (s % n):
        return s / n + 1;
    else:
        return s / n;
  
# Driver Code
n = 5;
s = 11;
print(int(minimumNumbers(n, s)));
  
# This code is contributed 
# by Shivi_Aggarwal


C#
// C# program to find the minimum numbers
// required to get to S
using System;
  
class GFG {
      
  
// Function to find the minimum
// numbers required to get to S
static int minimumNumbers(int n, int s)
{
    if ((s % n)>0)
        return s / n + 1;
    else
        return s / n;
}
  
// Driver Code
  
    public static void Main () {
        int n = 5;
    int s = 11;
    Console.WriteLine(minimumNumbers(n, s));
    }
}
// This code is contributed by shs..


PHP


输出:
3