求最小数 K，使得乘以 K 后的数组之和超过 S

原文:https://www . geeksforgeeks . org/find-最小数-k-so-k 乘 k 后的数组和-over-s/

给定一个由 N 元素组成的数组arr【】和一个整数 S ，任务是在将所有元素乘以 K 后，找到最小数量 K ，使得数组元素的总和不超过 S 。 示例:

输入: arr[] = { 1 }，S = 50 输出: 51 说明: 数组元素之和为 1。 现在 1 与 51 相乘得到 51，也就是> 50。 因此 K 的最小值为 51。 输入: arr[] = { 10，7，8，10，12，19 }，S = 200 输出: 4 说明: 数组元素之和为 66。 现在 66 与 4 相乘得到 256 > 200。 因此 K 的最小值为 4。

进场:

  • 求数组所有元素的和，存储在变量和中。
  • 取 上限 除以(S + 1)之和。这将是所需的最小值 k。

以下是上述方法的实现:

C++

// C++ implementation of the approach

#include <bits/stdc++.h>
using namespace std;

// Function to return the minimum value of k
// that satisfies the given condition
int findMinimumK(int a[], int n, int S)
{
    // store sum of array elements
    int sum = 0;

    // Calculate the sum after
    for (int i = 0; i < n; i++) {
        sum += a[i];
    }

    // return minimum possible K
    return ceil(((S + 1) * 1.0)
                / (sum * 1.0));
}

// Driver code
int main()
{
    int a[] = { 10, 7, 8, 10, 12, 19 };
    int n = sizeof(a) / sizeof(a[0]);
    int S = 200;

    cout << findMinimumK(a, n, S);

    return 0;
}

Java 语言(一种计算机语言，尤用于创建网站)

// Java implementation of the approach
import java.io.*;
import java.lang.Math;

class GFG {

// Function to return the minimum value of k
// that satisfies the given condition
static int findMinimumK(int a[], int n, int S)
{

    // Store sum of array elements
    int sum = 0;

    // Calculate the sum after
    for (int i = 0; i < n; i++)
    {
        sum += a[i];
    }

    // Return minimum possible K
    return (int) Math.ceil(((S + 1) * 1.0) /
                               (sum * 1.0));
}

// Driver code
public static void main(String[] args)
{
    int a[] = { 10, 7, 8, 10, 12, 19 };
    int n = a.length;
    int S = 200;
    System.out.print(findMinimumK(a, n, S));
}
}

// This code is contributed by shivanisinghss2110

Python 3

# Python3 implementation of the approach
import math

# Function to return the minimum value of k
# that satisfies the given condition
def findMinimumK(a, n, S) :

    # store sum of array elements
    sum = 0

    # Calculate the sum after
    for i in range(0,n):
        sum += a[i]

    # return minimum possible K
    return math.ceil(((S + 1) * 1.0)/(sum * 1.0))

# Driver code
a = [ 10, 7, 8, 10, 12, 19 ]
n = len(a)
s = 200
print(findMinimumK(a, n, s))

# This code is contributed by Sanjit_Prasad

C

// C# implementation of the approach
using System;

class GFG {

// Function to return the minimum value of k
// that satisfies the given condition
static int findMinimumK(int []a, int n, int S)
{

    // Store sum of array elements
    int sum = 0;

    // Calculate the sum after
    for(int i = 0; i < n; i++)
    {
       sum += a[i];
    }

    // Return minimum possible K
    return (int) Math.Ceiling(((S + 1) * 1.0) /
                                  (sum * 1.0));
}

// Driver code
public static void Main(String[] args)
{
    int []a = { 10, 7, 8, 10, 12, 19 };
    int n = a.Length;
    int S = 200;

    Console.Write(findMinimumK(a, n, S));
}
}

// This code is contributed by sapnasingh4991

java 描述语言

<script>

    // Javascript implementation of the approach

    // Function to return the minimum value of k
    // that satisfies the given condition
    function findMinimumK(a, n, S)
    {
        // store sum of array elements
        let sum = 0;

        // Calculate the sum after
        for (let i = 0; i < n; i++) {
            sum += a[i];
        }

        // return minimum possible K
        return Math.ceil(((S + 1) * 1.0) / (sum * 1.0));
    }

    let a = [ 10, 7, 8, 10, 12, 19 ];
    let n = a.length;
    let S = 200;

    document.write(findMinimumK(a, n, S));

</script>

Output: 

4

时间复杂度: O(N) 空间复杂度: O(1)

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