根据航空旅行的重量计算为行李支付的额外费用
给定一个大小为N的数组weight[] ，其中包含行李的重量。如果权重在W的阈值内，则不需要任何额外成本。但是在权重超过阈值后，他们需要根据下表支付额外的费用。任务是计算所需行李的额外费用。Range of exceeding weight Cost for the excess weight 0 – 50 100 51 – 100 200 101 – 150 300 151 – 200 500 > 200 1000
例子：
Input: weight[] = {5, 4, 3, 6}, W = 8
Output: 0
Explanation: No weight crosses the threshold. So no extra cost is incurred.
Input: weight[] = {120, 135, 280, 60, 300}, W = 90
Output: 1700
Explanation: The weight 120 is 30 more than the threshold. So, it requires 100 extra cost.
The weight 135 is 45 more than the threshold. So, it requires 100 extra cost.
The weight 280 is 190 more than the threshold. So, it requires 500 extra cost.
The weight 300 is 210 more than the threshold. So, it requires 1000 extra cost.
And the weight 60 is within threshold. So, it requires no cost.
The total extra cost is (100 + 100 + 500 + 1000) = 1700
方法：该方法基于以下观察。如果行李的重量超过阈值W ，则根据给定的表格会产生额外费用。请按照以下步骤解决问题：
- 从头开始迭代数组。
- 检查当前行李的重量是否超过W 。
- 如果有，则根据给定表格后的超重添加额外费用。
- 否则继续迭代。
- 检查当前行李的重量是否超过W 。
- 返回总额外费用
下面是上述方法的实现，
C++
// C++ code to implement the above approach
#include
using namespace std;
// Function to calculate the extra cost
int weighingMachine(int N, int weight[], int W)
{
int amount = 0;
// Loop to calculate the excess cost
for (int i = 0; i < N; i++) {
if (weight[i] - W > 0 && weight[i] - W <= 50)
amount += 100;
else if (weight[i] - W > 50 && weight[i] - W <= 100)
amount += 200;
else if (weight[i] - W > 100
&& weight[i] - W <= 150)
amount += 300;
else if (weight[i] - W > 150
&& weight[i] - W <= 200)
amount += 500;
else if (weight[i] - W > 200)
amount += 1000;
}
return amount;
}
// Driver code
int main()
{
int weight[] = { 120, 135, 280, 60, 300 };
int N = 5;
int W = 90;
cout << weighingMachine(N, weight, W);
}
// This code is contributed by Samim Hossain Mondal. Java
// Java code to implement the above approach
import java.util.*;
class GFG {
// Function to calculate the extra cost
static int weighingMachine(int N, int weight[], int W)
{
int amount = 0;
// Loop to calculate the excess cost
for (int i = 0; i < N; i++) {
if (weight[i] - W > 0
&& weight[i] - W <= 50)
amount += 100;
else if (weight[i] - W > 50
&& weight[i] - W <= 100)
amount += 200;
else if (weight[i] - W > 100
&& weight[i] - W <= 150)
amount += 300;
else if (weight[i] - W > 150
&& weight[i] - W <= 200)
amount += 500;
else if (weight[i] - W > 200)
amount += 1000;
}
return amount;
}
// Driver code
public static void main(String[] args)
{
int weight[] =
{ 120, 135, 280, 60, 300 };
int N = 5;
int W = 90;
System.out.println(
weighingMachine(N, weight, W));
}
}Python
# Python code to implement the above approach
# Function to calculate the extra cost
def weighingMachine(N, weight, W):
amount = 0;
# Loop to calculate the excess cost
for i in range(0, N):
if (weight[i] - W > 0 and weight[i] - W <= 50):
amount = amount + 100
elif (weight[i] - W > 50 and weight[i] - W <= 100):
amount = amount + 200
elif (weight[i] - W > 100
and weight[i] - W <= 150):
amount = amount + 300
elif (weight[i] - W > 150
and weight[i] - W <= 200):
amount = amount + 500;
elif (weight[i] - W > 200):
amount = amount + 1000
return amount
# Driver code
weight = [ 120, 135, 280, 60, 300 ]
N = 5
W = 90
print(weighingMachine(N, weight, W))
# This code is contributed by Samim Hossain Mondal.C#
// C# code to implement the above approach
using System;
public class GFG {
// Function to calculate the extra cost
static int weighingMachine(int N, int []weight, int W)
{
int amount = 0;
// Loop to calculate the excess cost
for (int i = 0; i < N; i++) {
if (weight[i] - W > 0
&& weight[i] - W <= 50)
amount += 100;
else if (weight[i] - W > 50
&& weight[i] - W <= 100)
amount += 200;
else if (weight[i] - W > 100
&& weight[i] - W <= 150)
amount += 300;
else if (weight[i] - W > 150
&& weight[i] - W <= 200)
amount += 500;
else if (weight[i] - W > 200)
amount += 1000;
}
return amount;
}
// Driver code
public static void Main(String[] args)
{
int []weight =
{ 120, 135, 280, 60, 300 };
int N = 5;
int W = 90;
Console.WriteLine(
weighingMachine(N, weight, W));
}
}
// This code is contributed by 29AjayKumarJavascript
1700时间复杂度： O(N)
辅助空间： O(1)