到达最后一站的最小总成本
给定一个数组，其中每个元素表示对应于每个站的巧克力数量，并且从站 I 移动到站 i+1，我们免费获得 A[I]–A[I+1]巧克力，如果这个数字是负数，我们也会失去那么多巧克力。 如果我们有非负数的巧克力，你只能从 I 站移动到 i+1 站。 给定一个巧克力 p 的成本，寻找从第一站(站 1)到达最后一站的最小成本的任务。 注:最初我们有 0 块巧克力。 示例:
Input : arr[] = {1, 2, 3} p = 10
Output : 30
Initial choc_buy = 1 (arr[0])
To reach index 0 to 1, arr[0] - arr[1] = -1,
so choc_buy +=1.
Similarly To reach index 2 to 1,
arr[1] - arr[2] = -1, so choc_buy +=1.
Total choc_buy = 3.
Total cost = choc_by * p = 3*10 = 30.
Input : arr[] = {10, 6, 11, 4, 7, 1} p = 5
Output : 55
Initial choc_buy = 10 (arr[0])
To reach index 0 to 1, arr[0] - arr[1] = 4,
so curr_choc =4.
Similarly To reach index 2 to 1,
arr[1] - arr[2] = -5, so curr_choc=4+-5 = -1.
choc_buy += abs(-1).
So, similarly till last station, Total choc_buy = 11.
Total cost = choc_by * p = 11*5 = 55.
进场: 1。用数组的第一个元素初始化 choc_buy，curr_choc =0。 2。将 arr[i]-arr[i+1]添加到每个 I 的 curr_choc 中。 a)检查 curr_choc 是否为负数。 choc _ buy+= ABS(arr[I]-arr[I+1]) curr _ choc = 0 3。返回 choc_buy * p。
C++
// C++ program to calculate
// minimum cost for candies
#include <bits/stdc++.h>
using namespace std;
// Function to find minimum cost
// to be incurred
int findMinCost(int arr[], int n, int choc_cost) {
// To reach first station, initial
// chocolates required
int choc_buy = arr[0];
int curr_choc = 0;
// Start traversing
for (int i = 0; i < n - 1; i++) {
// Find no. of chocolates
// lose or gain
int choc = arr[i] - arr[i + 1];
// Add into curr_coc
curr_choc += choc;
// if no. of chocolates becomes
// negative that means we have
// to buy that no. of chocolates
if (curr_choc < 0) {
choc_buy += abs(curr_choc);
curr_choc = 0;
}
}
// Return cost required
return choc_buy * choc_cost;
}
// Drivers code
int main() {
int arr[] = {10, 6, 11, 4, 7, 1};
int n = sizeof(arr) / sizeof(arr[0]);
// Price of each candy
int p = 5;
cout << findMinCost(arr, n, p);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to calculate
// minimum cost for candies
import java.io.*;
class GFG
{
// Function to find minimum cost
// to be incurred
static int findMinCost(int arr[], int n, int choc_cost)
{
// To reach first station, initial
// chocolates required
int choc_buy = arr[0];
int curr_choc = 0;
// Start traversing
for (int i = 0; i < n - 1; i++)
{
// Find no. of chocolates
// lose or gain
int choc = arr[i] - arr[i + 1];
// Add into curr_coc
curr_choc += choc;
// if no. of chocolates becomes
// negative that means we have
// to buy that no. of chocolates
if (curr_choc < 0)
{
choc_buy += Math.abs(curr_choc);
curr_choc = 0;
}
}
// Return cost required
return choc_buy * choc_cost;
}
// Drivers code
public static void main (String[] args)
{
int arr[] = {10, 6, 11, 4, 7, 1};
int n = arr.length;
// Price of each candy
int p = 5;
System.out.println ( findMinCost(arr, n, p));
}
}
// This code is contributed by vt_m
Python 3
# Python 3 program to calculate
# minimum cost for candies
# Function to find minimum cost
# to be incurred
def findMinCost(arr, n, choc_cost):
# To reach first station,
# initial chocolates required
choc_buy = arr[0]
curr_choc = 0
# Start traversing
for i in range(0,n - 1):
# Find no. of chocolates lose
# or gain
choc = arr[i] - arr[i + 1]
# Add into curr_coc
curr_choc += choc
# if no. of chocolates becomes
# negative that means we have
# to buy that no. of chocolates
if (curr_choc < 0):
choc_buy += abs(curr_choc)
curr_choc = 0
# Return cost required
return choc_buy * choc_cost
# Drivers code
arr = [10, 6, 11, 4, 7, 1]
n = len(arr)
# Price of each candy
p = 5
print(findMinCost(arr, n, p))
# This code is contributed by Smitha Dinesh Semwal
C
// C# program to calculate
// minimum cost for candies
using System;
class GFG
{
// Function to find minimum cost
// to be incurred
static int findMinCost(int []arr,
int n, int choc_cost)
{
// To reach first station, initial
// chocolates required
int choc_buy = arr[0];
int curr_choc = 0;
// Start traversing
for (int i = 0; i < n - 1; i++)
{
// Find no. of chocolates
// lose or gain
int choc = arr[i] - arr[i + 1];
// Add into curr_coc
curr_choc += choc;
// if no. of chocolates becomes
// negative that means we have
// to buy that no. of chocolates
if (curr_choc < 0)
{
choc_buy += Math.Abs(curr_choc);
curr_choc = 0;
}
}
// Return cost required
return choc_buy * choc_cost;
}
// Drivers code
public static void Main ()
{
int []arr = {10, 6, 11, 4, 7, 1};
int n = arr.Length;
// Price of each candy
int p = 5;
Console.WriteLine( findMinCost(arr, n, p));
}
}
// This code is contributed by vt_m
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to calculate
// minimum cost for candies
// Function to find minimum cost
// to be incurred
function findMinCost($arr, $n, $choc_cost)
{
// To reach first station, initial
// chocolates required
$choc_buy = $arr[0];
$curr_choc = 0;
// Start traversing
for ($i = 0; $i < $n - 1; $i++) {
// Find no. of chocolates
// lose or gain
$choc = $arr[$i] - $arr[$i + 1];
// Add into curr_coc
$curr_choc += $choc;
// if no. of chocolates becomes
// negative that means we have
// to buy that no. of chocolates
if ($curr_choc < 0) {
$choc_buy += abs($curr_choc);
$curr_choc = 0;
}
}
// Return cost required
return $choc_buy * $choc_cost;
}
// Driver Code
$arr = array(10, 6, 11, 4, 7, 1);
$n = count($arr);
// Price of each candy
$p = 5;
echo findMinCost($arr, $n, $p);
// This code is contributed by Sam007
?>
java 描述语言
<script>
// Javascript program to calculate
// minimum cost for candies
// Function to find minimum cost
// to be incurred
function findMinCost(arr, n, choc_cost) {
// To reach first station, initial
// chocolates required
var choc_buy = arr[0];
var curr_choc = 0;
// Start traversing
for (var i = 0; i < n - 1; i++) {
// Find no. of chocolates
// lose or gain
var choc = arr[i] - arr[i + 1];
// Add into curr_coc
curr_choc += choc;
// if no. of chocolates becomes
// negative that means we have
// to buy that no. of chocolates
if (curr_choc < 0) {
choc_buy += Math.abs(curr_choc);
curr_choc = 0;
}
}
// Return cost required
return (choc_buy * choc_cost);
}
var arr = [10, 6, 11, 4, 7, 1];
var n = 6;
// Price of each candy
var p = 5;
document.write(findMinCost(arr, n, p));
</script>
Output:
55
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