根据给定条件计算 N 天后放在盒子里的钱

给定 7 个空盒子b1、b2、b3、b4、b5、b6、b7和一个整数N ，任务是根据以下条件找出N天后可以放入盒子中的总金额：

1. 每天，钱只能以循环方式放入一个盒子b1、b2、b3、b4、b5、b6、b7、b1、b2、......等等。
2. 在方框b1中，比方框b1中已经存在的钱多放1 。
3. 在除b1之外的每个盒子里，比前一个盒子里的钱多放1 。

例子：

Input: N = 4
Output: 15
Explanation:
Putting money in the box b1 on day 1 = 1
Putting money in the box b2 on day 2 = 2
Putting money in the box b3 on day 3 = 3
Putting money in the box b4 on day 4 = 4
Putting money in the box b5 on day 5 = 5
After the 5th day, total amount = 1 + 2 + 3 + 4 + 5 = 15

Input: N = 15
Output: 66
Explanation: After the 15th day, the total amount = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 3 = 66

方法：按照以下步骤解决问题

1. 第i^天花的钱是((i – 1)/ 7) + ((i – 1) % 7 + 1)，其中i在[1, N]范围内
2. 模拟相同的天数[1, N]
3. 打印总成本。

下面是上述方法的实现：

C++

// C++ program for
// the above approach
#include 
using namespace std;
 
// Function to find the total money
// placed in boxes after N days
int totalMoney(int N)
{
     
    // Stores the total money
    int ans = 0;
 
    // Iterate for N days
    for(int i = 0; i < N; i++)
    {
         
        // Adding the Week number
        ans += i / 7;
 
        // Adding previous amount + 1
        ans += (i % 7 + 1);
    }
 
    // Return the total amount
    return ans;
}
 
// Driver code
int main()
{  
     
    // Input
    int N = 15;
 
    // Function call to find
    // total money placed
    cout << totalMoney(N);
}
 
// This code is contributed khushboogoyal499

Java

// Java program for
// the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to find the total money
    // placed in boxes after N days
    public static int totalMoney(int N)
    {
 
        // Stores the total money
        int ans = 0;
 
        // Iterate for N days
        for (int i = 0; i < N; i++) {
 
            // Adding the Week number
            ans += i / 7;
 
            // Adding previous amount + 1
            ans += (i % 7 + 1);
        }
 
        // Return the total amount
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Input
        int N = 15;
 
        // Function call to find
        // total money placed
        System.out.println(
            totalMoney(N));
    }
}

Python

# Python program for
# the above approach
 
# Function to find the total money
# placed in boxes after N days
def totalMoney(N):
     
    # Stores the total money
    ans = 0
 
    # Iterate for N days
    for i in range(0, N):
      
        # Adding the Week number
        ans += i / 7
 
        # Adding previous amount + 1
        ans += (i % 7 + 1)
      
    # Return the total amount
    return ans
  
# Driver code
 
# Input
N = 15
 
# Function call to find
# total money placed
print(totalMoney(N))
 
# This code is contributed by shivanisinghss2110

C#

// C# program for
// the above approach
using System;
 
class GFG{
     
// Function to find the total money
// placed in boxes after N days
public static int totalMoney(int N)
{
     
    // Stores the total money
    int ans = 0;
 
    // Iterate for N days
    for(int i = 0; i < N; i++)
    {
         
        // Adding the Week number
        ans += i / 7;
 
        // Adding previous amount + 1
        ans += (i % 7 + 1);
    }
 
    // Return the total amount
    return ans;
}
 
// Driver code
static public void Main()
{
     
    // Input
    int N = 15;
 
    // Function call to find
    // total money placed
    Console.WriteLine(totalMoney(N));
}
}
 
// This code is contributed by offbeat

Javascript

 
// JavaScript Program to implement
// the above approach
 
    // Function to find the total money
    // placed in boxes after N days
    function totalMoney(N)
    {
 
        // Stores the total money
        let ans = 0;
 
        // Iterate for N days
        for (let i = 0; i < N; i++) {
 
            // Adding the Week number
            ans += Math.floor(i / 7);
 
            // Adding previous amount + 1
            ans += (i % 7 + 1);
        }
 
        // Return the total amount
        return ans;
    }
 
 
// Driver Code
 
    // Input
        let N = 15;
 
        // Function call to find
        // total money placed
       document.write(
            totalMoney(N));
 

输出：

66

时间复杂度： O(N)
辅助空间： O(1)

高效方法：上述方法可以通过找出已完成的周数和上周剩余的天数来优化。

请按照以下步骤解决问题：

1. 初始化变量X和Y ，分别存储可以在完整周和部分周中放置的金额。
2. 每周的钱可以这样计算：
   • ^第一周：1 2 3 4 5 6 7 = 28 + (7 x 0)
   • 第 2^周：2 3 4 5 6 7 8 = 28 + (7 x 1)
   • ^第三周：3 4 5 6 7 8 9 = 28 + (7 x 2)
   • 第 4^周：4 5 6 7 8 9 10 = 28 + (7 x 3) 等等。
3. 因此，更新：
   X = 28 + 7 x（完成周数 - 1）
   Y =剩余天数总和 +（完整周数 * 上周剩余天数）
4. 因此，总量等于X + Y 。打印放置的总金额。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find total
// money placed in the box
int totalMoney(int N)
{
    // Number of complete weeks
    int CompWeeks = N / 7;
 
    // Remaining days in
    // the last week
    int RemDays = N % 7;
 
    int X = 28 * CompWeeks
            + 7 * (CompWeeks
                   * (CompWeeks - 1) / 2);
 
    int Y = RemDays
                * (RemDays + 1) / 2
            + CompWeeks * RemDays;
 
    int cost = X + Y;
 
    cout << cost << '\n';
}
 
// Driver Code
int main()
{
    // Input
    int N = 15;
 
    // Function call to find
    // the total money placed
    totalMoney(N);
 
    return 0;
}

Java

// Java program for above approach
 
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
// Function to find total
// money placed in the box
static void totalMoney(int N)
{
    // Number of complete weeks
    int CompWeeks = N / 7;
   
    // Remaining days in
    // the last week
    int RemDays = N % 7;
   
    int X = 28 * CompWeeks
            + 7 * (CompWeeks
                   * (CompWeeks - 1) / 2);
   
    int Y = RemDays
                * (RemDays + 1) / 2
            + CompWeeks * RemDays;
   
    int cost = X + Y;
   
    System.out.print(cost);
}
 
 
    // Driver Code
    public static void main(String[] args)
    {
    // Input
    int N = 15;
   
    // Function call to find
    // the total money placed
    totalMoney(N);
    }
}
 
// This code is contributed by souravghosh0416.

C＃

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find total
// money placed in the box
static void totalMoney(int N)
{
     
    // Number of complete weeks
    int CompWeeks = N / 7;
    
    // Remaining days in
    // the last week
    int RemDays = N % 7;
    
    int X = 28 * CompWeeks + 7 *
    (CompWeeks * (CompWeeks - 1) / 2);
    
    int Y = RemDays * (RemDays + 1) / 2 +
          CompWeeks * RemDays;
    
    int cost = X + Y;
    
    Console.WriteLine(cost);
}   
 
// Driver Code
public static void Main()
{
     
    // Input
    int N = 15;
    
    // Function call to find
    // the total money placed
    totalMoney(N);
}
}
 
// This code is contriobuted by sanjoy_62

Javascript

 
// JavaScript Program to implement
// the above approach
 
// Function to find total
// money placed in the box
function totalMoney( N)
{
    // Number of complete weeks
    let CompWeeks = Math.floor(N / 7);
 
    // Remaining days in
    // the last week
    let RemDays = N % 7;
 
    let X = 28 * CompWeeks
            + 7 * Math.floor((CompWeeks
                   * (CompWeeks - 1) / 2));
 
    let Y = RemDays
                * Math.floor((RemDays + 1) / 2)
            + CompWeeks * RemDays;
 
    let cost = X + Y;
 
    document.write(cost ,'<br>');
}
 
// Driver Code
 
    // Input
    let N = 15;
 
    // Function call to find
    // the total money placed
    totalMoney(N);
 

输出：

66

时间复杂度： O(1)
辅助空间： O(1)