给定的正整数和一个值K的阵列，该任务是清空在阵列小于或等于k小阵列，使得每个小阵只能包含在从给定阵列的单个时隙/索引最大P个元素。找出P的最小值。
例子：
Input: arr[] = {1, 2, 3, 4, 5}, K = 7
Output: 3
Explanation:
We put 1 into the first small array,
2 into the second array,
3 into the third array,
After this, we divide the other elements as 1 + 3 and 2 + 3
These 4 elements can be put into the remaining 4 boxes.
So, the required value of P is 3.
Input: arr[] = {23, 1, 43, 66, 220}, K = 102
Output: 4
方法：要解决此问题，我们需要二进制搜索答案。
- 首先，我们将下限设置为1，将上限设置为给定数组的最大值。
- 现在，我们可以在此范围内执行二进制搜索。对于特定的容量值，我们计算需要包含项目中所有值的小型阵列的数量。
- 如果所需的小数组数量大于K，则答案肯定会更大，因此，我们会修剪搜索的左侧。如果它小于或等于K，我们将该值保留为可能的答案，并修剪搜索的右侧。
下面是上述方法的实现。
C++
// C++ program to find the Minimum
// capacity of small arrays needed
// to contain all element of
// the given array
#include
using namespace std;
// Function returns the value
// of Minimum capacity needed
int MinimumCapacity(vector arr,
int K)
{
// Initializing maximum
// value
int maxVal = arr[0];
// Finding maximum value
// in arr
for (auto x : arr)
maxVal = max(maxVal, x);
int l = 1, r = maxVal, m;
int ans, req;
// Binary Search the answer
while (l <= r)
{
// m is the mid-point
// of the range
m = l + (r - l) / 2;
req = 0;
// Finding the total number of
// arrays needed to completely
// contain the items array
// with P = req
for (auto x : arr)
req += x / m + (x % m > 0);
// If the required number of
// arrays is more than K, it
// means we need to increase
// the value of P
if (req > K)
l = m + 1;
else
// If the required number of
// arrays is less than or equal
// to K, it means this is a
// possible answer and we go to
// check if any smaller possible
// value exists for P
ans = m, r = m - 1;
}
return ans;
}
// Driver Code
int main()
{
// Given array
vector arr = { 1, 2, 3, 4, 5 };
// Number of available small arrays
int K = 7;
cout << MinimumCapacity(arr, K);
return 0;
} Java
// Java program to find the Minimum
// capacity of small arrays needed
// to contain all element of
// the given array
class GFG{
// Function returns the value
// of Minimum capacity needed
static int MinimumCapacity(int []arr,
int K)
{
// Initializing maximum
// value
int maxVal = arr[0];
// Finding maximum value
// in arr
for (int x : arr)
maxVal = Math.max(maxVal, x);
int l = 1, r = maxVal, m;
int ans = 0, req;
// Binary Search the answer
while (l <= r)
{
// m is the mid-point
// of the range
m = l + (r - l) / 2;
req = 0;
// Finding the total number of
// arrays needed to completely
// contain the items array
// with P = req
for (int x : arr)
req += x / m + (x % m > 0 ? 1 : 0);
// If the required number of
// arrays is more than K, it
// means we need to increase
// the value of P
if (req > K)
l = m + 1;
else
{
// If the required number of
// arrays is less than or equal
// to K, it means this is a
// possible answer and we go to
// check if any smaller possible
// value exists for P
ans = m;
r = m - 1;
}
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
// Given array
int []arr = { 1, 2, 3, 4, 5 };
// Number of available small arrays
int K = 7;
System.out.print(MinimumCapacity(arr, K));
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find the Minimum
# capacity of small arrays needed
# to contain all element of
# the given array
# Function returns the value
# of Minimum capacity needed
def MinimumCapacity(arr, K):
# Initializing maximum
# value
maxVal = arr[0]
# Finding maximum value
# in arr
for x in arr:
maxVal = max(maxVal, x)
l = 1
r = maxVal
m = 0
ans = 0
req = 0
# Binary Search the answer
while l <= r:
# m is the mid-point
# of the range
m = l + (r - l) // 2
req = 0
# Finding the total number of
# arrays needed to completely
# contain the items array
# with P = req
for x in arr:
req += x // m + (x % m > 0)
# If the required number of
# arrays is more than K, it
# means we need to increase
# the value of P
if req > K:
l = m + 1
else:
# If the required number of
# arrays is less than or equal
# to K, it means this is a
# possible answer and we go to
# check if any smaller possible
# value exists for P
ans = m
r = m - 1
return ans
#Driver Code
# Given array
arr = [ 1, 2, 3, 4, 5 ]
# Number of available small arrays
K = 7
print(MinimumCapacity(arr, K))
# This code is contributed by divyamohan123C#
// C# program to find the minimum
// capacity of small arrays needed
// to contain all element of
// the given array
using System;
class GFG{
// Function returns the value
// of minimum capacity needed
static int MinimumCapacity(int []arr,
int K)
{
// Initializing maximum
// value
int maxVal = arr[0];
// Finding maximum value
// in arr
foreach (int x in arr)
maxVal = Math.Max(maxVal, x);
int l = 1, r = maxVal, m;
int ans = 0, req;
// Binary Search the answer
while (l <= r)
{
// m is the mid-point
// of the range
m = l + (r - l) / 2;
req = 0;
// Finding the total number of
// arrays needed to completely
// contain the items array
// with P = req
foreach (int x in arr)
req += x / m + (x % m > 0 ? 1 : 0);
// If the required number of
// arrays is more than K, it
// means we need to increase
// the value of P
if (req > K)
l = m + 1;
else
{
// If the required number of
// arrays is less than or equal
// to K, it means this is a
// possible answer and we go to
// check if any smaller possible
// value exists for P
ans = m;
r = m - 1;
}
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = { 1, 2, 3, 4, 5 };
// Number of available small arrays
int K = 7;
Console.Write(MinimumCapacity(arr, K));
}
}
// This code is contributed by 29AjayKumar输出：
3