解决生活问题

你在俄罗斯萨马拉工作了几天。每一天都有一个新的每单位工作工资和一个新的每单位食物成本。工作 1 单位消耗 1 单位能量，吃 1 单位食物增加 1 单位能量。以下是您的工作的一些规格：

• 你到达时没有钱，但有能量。
• 你的能量永远不会超过你到达时的能量，它永远不会是消极的。
• 您每天可以做任何数量的工作(可能根本不做任何工作)，仅受您的精力限制。
• 当你的能量为零时，你无法工作。您每天可以吃任意数量的食物(可能根本没有任何食物)，这受您拥有的金钱的限制。
• 当你的钱为零时，你不能吃饭。您可以在一天结束时进食，进食后不能返回工作。
• 第二天就可以上班了。

你的真正目标是带着尽可能多的钱回家。计算您可以带回家的最大金额。

例子：

Input: N = 3, Earning=[1, 2, 4], Cost[]=[1, 3, 6], E = 5
Output: 20
Explanation:
Day 1: 1 unit work is worth 1, and 1 unit food costs 1. There is no financial incentive to go to work this day.
Day 2: 1 unit work earns 2, and 1 unit food costs 3. Thus, you spend more to eat then the total earning so there is no financial incentive to go to work on this day.
Day 3: You earn 4 units per unit of work. The cost of food is irrelevant this day, as you are leaving for the home straight from work.
You spend all of your energy working, the pay is 5 × 4 = 20 units of money and go home without buying dinner.

Input: N=2, Earning=[1, 2], Cost=[1, 4], E=5
Output: Total Profit = 0 + 10 = 10
Explanation: 
First Day: Skip
Second Day: 5*2=10

方法：方法是遍历给定的收益数组和成本数组，计算每一天的净利润。以下是步骤：

1. 对于earning []和cost[] 中的每个元素，将earning[i]与cost[i] 进行比较。
2. 如果Earning[i] 小于或等于 cost[i] ，则跳过当天的工作，因为费用成本大于收入。因此，根本没有盈利。
3. 如果收入[i] 大于成本[i] ，则通过将当天的收入乘以总能量单位来计算总收入，然后减去当天的食物成本和能量单位的乘积来计算利润那天。
4. 如果有最后一个工作日，那么通过将当天的收入乘以总能量单位来计算总收入，这将是您当天的利润，因为工作不再需要能量。
5. 完成上述步骤后打印总利润。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that calculates the profit
// with the earning and cost of expenses
int calculateProfit(int n, int* earnings,
                    int* cost, int e)
{
    // To store the total Profit
    int profit = 0;
 
    // Loop for n number of days
    for (int i = 0; i < n; i++) {
 
        int earning_per_day = 0;
        int daily_spent_food = 0;
 
        // If last day, no need to buy food
        if (i == (n - 1)) {
            earning_per_day = earnings[i] * e;
            profit = profit + earning_per_day;
            break;
        }
 
        // Else buy food to gain
        // energy for next day
        if (cost[i] < earnings[i]) {
 
            // Update earning per day
            earning_per_day = earnings[i] * e;
            daily_spent_food = cost[i] * e;
 
            // Update profit with daily spent
            profit = profit + earning_per_day
                     - daily_spent_food;
        }
    }
 
    // Print the profit
    cout << profit << endl;
}
 
// Driver Code
int main()
{
    // Given days
    int n = 4;
 
    // Given earnings
    int earnings[] = { 1, 8, 6, 7 };
 
    // Given cost
    int cost[] = { 1, 3, 4, 1 };
 
    // Given energy e
    int e = 5;
 
    // Function Call
    calculateProfit(n, earnings, cost, e);
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function that calculates the profit
// with the earning and cost of expenses
static void calculateProfit(int n, int []earnings,
                            int []cost, int e)
{
     
    // To store the total Profit
    int profit = 0;
 
    // Loop for n number of days
    for(int i = 0; i < n; i++)
    {
        int earning_per_day = 0;
        int daily_spent_food = 0;
 
        // If last day, no need to buy food
        if (i == (n - 1))
        {
            earning_per_day = earnings[i] * e;
            profit = profit + earning_per_day;
            break;
        }
 
        // Else buy food to gain
        // energy for next day
        if (cost[i] < earnings[i])
        {
 
            // Update earning per day
            earning_per_day = earnings[i] * e;
            daily_spent_food = cost[i] * e;
 
            // Update profit with daily spent
            profit = profit + earning_per_day -
                              daily_spent_food;
        }
    }
 
    // Print the profit
    System.out.print(profit + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given days
    int n = 4;
 
    // Given earnings
    int earnings[] = { 1, 8, 6, 7 };
 
    // Given cost
    int cost[] = { 1, 3, 4, 1 };
 
    // Given energy e
    int e = 5;
 
    // Function call
    calculateProfit(n, earnings, cost, e);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
 
# Function that calculates the profit
# with the earning and cost of expenses
def calculateProfit(n, earnings, cost, e):
     
    # To store the total Profit
    profit = 0;
 
    # Loop for n number of days
    for i in range(n):
        earning_per_day = 0;
        daily_spent_food = 0;
 
        # If last day, no need to buy food
        if (i == (n - 1)):
            earning_per_day = earnings[i] * e;
            profit = profit + earning_per_day;
            break;
 
        # Else buy food to gain
        # energy for next day
        if (cost[i] < earnings[i]):
             
            # Update earning per day
            earning_per_day = earnings[i] * e;
            daily_spent_food = cost[i] * e;
 
            # Update profit with daily spent
            profit = (profit + earning_per_day -
                               daily_spent_food);
 
    # Print the profit
    print(profit);
 
# Driver Code
if __name__ == '__main__':
     
    # Given days
    n = 4;
 
    # Given earnings
    earnings = [ 1, 8, 6, 7 ];
 
    # Given cost
    cost = [ 1, 3, 4, 1 ];
 
    # Given energy e
    e = 5;
 
    # Function call
    calculateProfit(n, earnings, cost, e);
 
# This code is contributed by PrinciRaj1992

C#

// C# program for the above approach
using System;
class GFG{
 
// Function that calculates the profit
// with the earning and cost of expenses
static void calculateProfit(int n, int []earnings,
                            int []cost, int e)
{
     
    // To store the total Profit
    int profit = 0;
 
    // Loop for n number of days
    for(int i = 0; i < n; i++)
    {
        int earning_per_day = 0;
        int daily_spent_food = 0;
 
        // If last day, no need to buy food
        if (i == (n - 1))
        {
            earning_per_day = earnings[i] * e;
            profit = profit + earning_per_day;
            break;
        }
 
        // Else buy food to gain
        // energy for next day
        if (cost[i] < earnings[i])
        {
 
            // Update earning per day
            earning_per_day = earnings[i] * e;
            daily_spent_food = cost[i] * e;
 
            // Update profit with daily spent
            profit = profit + earning_per_day -
                              daily_spent_food;
        }
    }
 
    // Print the profit
    Console.Write(profit + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given days
    int n = 4;
 
    // Given earnings
    int []earnings = { 1, 8, 6, 7 };
 
    // Given cost
    int []cost = { 1, 3, 4, 1 };
 
    // Given energy e
    int e = 5;
 
    // Function call
    calculateProfit(n, earnings, cost, e);
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// JavaScript program for the above approach
 
// Function that calculates the profit
// with the earning and cost of expenses
function calculateProfit(n, earnings, cost, e)
{
       
    // To store the total Profit
    let profit = 0;
   
    // Loop for n number of days
    for(let i = 0; i < n; i++)
    {
        let earning_per_day = 0;
        let daily_spent_food = 0;
   
        // If last day, no need to buy food
        if (i == (n - 1))
        {
            earning_per_day = earnings[i] * e;
            profit = profit + earning_per_day;
            break;
        }
   
        // Else buy food to gain
        // energy for next day
        if (cost[i] < earnings[i])
        {
   
            // Update earning per day
            earning_per_day = earnings[i] * e;
            daily_spent_food = cost[i] * e;
   
            // Update profit with daily spent
            profit = profit + earning_per_day -
                              daily_spent_food;
        }
    }
   
    // Prlet the profit
    document.write(profit );
}
 
// Driver Code
 
     // Given days
    let n = 4;
   
    // Given earnings
    let earnings = [ 1, 8, 6, 7 ];
   
    // Given cost
    let cost = [ 1, 3, 4, 1 ];
   
    // Given energy e
    let e = 5;
   
    // Function call
    calculateProfit(n, earnings, cost, e);
      

输出：

70

时间复杂度： O(N)，其中 N 是天数
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live