给定一个由N 个元素组成的数组arr[]和一个整数S ,任务是找到最小数K ,使得在将所有元素乘以K后,数组元素的总和不超过S。
例子:
Input: arr[] = { 1 }, S = 50
Output: 51
Explanation:
The sum of array elements is 1.
Now the multiplication of 1 with 51 gives 51 which is > 50.
Hence the minimum value of K is 51.
Input: arr[] = { 10, 7, 8, 10, 12, 19 }, S = 200
Output: 4
Explanation:
The sum of array elements is 66.
Now the multiplication of 66 with 4 gives 256 > 200.
Hence the minimum value of K is 4.
方法:
- 查找数组所有元素的总和,将其存储在变量sum 中。
- 将 (S + 1) 的ceil除法与 sum 相加。这将是所需的 K 最小值。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
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
// 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
Python3
# 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
Javascript
输出:
4
时间复杂度: O(N)
空间复杂度: O(1)
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