当一个物品总成本的其他物品免费时获得最大物品
原文:https://www . geeksforgeeks . org/get-最大项目数-当一个项目的总成本中的其他项目是免费的/
给定“N”项的价格列表。一个人只能购买任何价格的一件物品，他可以免费获得其他物品，这样其余物品的总价就不会超过购买物品的价格。任务是找到这个人能拥有的最大物品数量。 例:
输入: n = 5，arr = {5，3，1，5，6} 输出: 3 该人可以购买价格为 5 或 6 的任何物品，并免费下载价格为 1 和 3 的物品。所以，他最多只能得到 3 件物品。 输入: n = 2，arr = {7，7} 输出: 2
进场:
这个人应该购买最贵的物品，然后从最便宜的价格开始购买物品(直到总价小于或等于购买的物品)，以使物品的总数最大化。 因此，我们对价格列表进行排序并选择最后一个元素，然后我们将从第一个索引迭代到第 n-2 个索引，并检查总和是否小于或等于最后一个元素。
以下是上述方法的实现:
C++
// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count the
// total number of items
int items(int n, int a[]){
// Sort the prices
sort(a,a+n);
// Choose the last element
int z = a[n-1];
// Initial count of item
int x = 1;
// Sum to keep track of
// the total price
// of free items
int s = 0;
for (int i=0;i<n-1;i++)
{
s += a[i];
// If total is less than
// or equal to z then
// we will add 1 to the answer
if (s <= z)
x+= 1;
else
break;
}
return x;
}
int main()
{
int n = 5;
int a[]= {5, 3, 1, 5, 6};
cout<<items(n, a);
}
//contributed by Arnab Kundu
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of
// the above approach
import java.util.Arrays;
import java.io.*;
class GFG {
// Function to count the
// total number of items
static int items(int n, int a[]){
// Sort the prices
Arrays.sort(a);
// Choose the last element
int z = a[n-1];
// Initial count of item
int x = 1;
// Sum to keep track of
// the total price
// of free items
int s = 0;
for (int i=0;i<n-1;i++)
{
s += a[i];
// If total is less than
// or equal to z then
// we will add 1 to the answer
if (s <= z)
x+= 1;
else
break;
}
return x;
}
// Driver code
public static void main (String[] args) {
int n = 5;
int a[]= {5, 3, 1, 5, 6};
System.out.println(items(n, a));
}
//This code is contributed by ajit
}
Python 3
# Python3 implementation of
# the above approach
# Function to count the
# total number of items
def items(n, a):
# Sort the prices
a.sort()
# Choose the last element
z = a[n-1]
# Initial count of item
x = 1
# Sum to keep track of
# the total price
# of free items
s = 0
for i in range(0, n-1):
s += a[i]
# If total is less than
# or equal to z then
# we will add 1 to the answer
if (s <= z):
x+= 1
else:
break
return x
n = 5
a = [5, 3, 1, 5, 6]
print(items(n, a))
C
// C# implementation of the
// above approach
using System;
class GFG
{
// Function to count the
// total number of items
static int items(int n, int []a)
{
// Sort the prices
Array.Sort(a);
// Choose the last element
int z = a[n - 1];
// Initial count of item
int x = 1;
// Sum to keep track of
// the total price
// of free items
int s = 0;
for (int i = 0; i < n - 1; i++)
{
s += a[i];
// If total is less than or equal to z
// then we will add 1 to the answer
if (s <= z)
x += 1;
else
break;
}
return x;
}
// Driver code
static public void Main ()
{
int n = 5;
int []a = {5, 3, 1, 5, 6};
Console.WriteLine(items(n, a));
}
}
// This code is contributed
// by akt_mit
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
//PHP implementation of
// the above approach
// Function to count the
// total number of items
function items($n, $a){
// Sort the prices
sort($a);
// Choose the last element
$z = $a[$n-1];
// Initial count of item
$x = 1;
// Sum to keep track of
// the total price
// of free items
$s = 0;
for ($i=0;$i<$n-1;$i++)
{
$s += $a[$i];
// If total is less than
// or equal to z then
// we will add 1 to the answer
if ($s <= $z)
$x+= 1;
else
break;
}
return $x;
}
//Code driven
$n = 5;
$a= array(5, 3, 1, 5, 6);
echo items($n, $a);
//This code is contributed by ajit
?>
java 描述语言
<script>
// Javascript implementation of the above approach
// Function to count the
// total number of items
function items(n, a)
{
// Sort the prices
a.sort(function(a, b){return a - b});
// Choose the last element
let z = a[n - 1];
// Initial count of item
let x = 1;
// Sum to keep track of
// the total price
// of free items
let s = 0;
for (let i = 0; i < n - 1; i++)
{
s += a[i];
// If total is less than or equal to z
// then we will add 1 to the answer
if (s <= z)
x += 1;
else
break;
}
return x;
}
let n = 5;
let a = [5, 3, 1, 5, 6];
document.write(items(n, a));
</script>
Output:
3
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