规定时间内可装满的容器数

给定一个数字N和一个时间X单位，如果容器按金字塔方式排列，则任务是找到在 X 单位中完全装满的容器数量，如下所示。

注意：下面的例子是 N = 3 的金字塔排列，其中N表示容器的金字塔形排列中的级别数，这样级别 1 有 1 个容器，级别 2 有 2 个容器，直到级别 N 。液体总是倒在第一层最上面的容器中。当一层容器从两侧溢出时，下层容器就会被装满。每秒倒入的液体量等于容器的体积。

例子：

Input: N = 3, X = 5
Output: 4
After 1 second, the container at level 1 gets fully filled.
After 2 seconds, the 2 containers at level 2 are half-filled.
After 3 seconds, the 2 containers at level 2 are fully filled.
After 4 seconds, out of the 3 containers at level 3, the 2 containers at the ends are quarter-filled and the container at the center is half-filled.
After 5 seconds, out of the 3 containers at level 3, the 2 containers at the ends are half-filled and the container at the center is fully filled.

Input: N = 4, X = 8
Output: 6

编程需要懂一点英语

方法：这个问题是用贪心方法解决的。以下是步骤：

容器装满后，液体从容器的两侧等量溢出并充满其下方的容器。
将此容器金字塔视为一个矩阵。因此， cont[i][j]是^第i行^第j^个容器中的液体。
因此，当发生溢出时，液体流向cont[i+1][j]和cont[i+1][j+1] 。
由于液体将倾倒X秒，因此倾倒的液体总量为X 个单位。
因此，可以倒入容器的最大值可以为X 个单位，而可以倒入以完全填充容器的最小值为1 个单位。

由于液体总是倒入最上面的容器，让最上面的容器有一个最大值，即 X 个单位。
算法步骤如下：

对于每个级别的n 个容器，从1 到 N开始外循环。在这个循环中，为每行的每个容器启动另一个循环，从1 到 i 。还要声明一个计数器， count = 0 ，用于计算正在装满的容器。
如果cont[i][j]的值大于或等于1 （表示已填充），则 count 增加1 ，然后在cont[i+1][j] & g[i+1 ][j+1]液体被倾倒并且它们的值分别增加了(g[i][j]-1)/2 的值，因为液体被分成了一半。
以这种方式继续循环，并增加每个填充容器的计数。当循环结束时，我们的计数将是所需的答案。

下面是上述方法的实现：

C++

// C++ program for the above problem
#include 
using namespace std;
 
// matrix of containers
double cont[1000][1000];
 
// function to find the number
// of containers that will be
// filled in X seconds
void num_of_containers(int n,
                       double x)
{
    int count = 0;
 
    // container on top level
    cont[1][1] = x;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
 
            // if container gets filled
            if (cont[i][j] >= (double)1) {
                count++;
 
                // dividing the liquid
                // equally in two halves
                cont[i + 1][j]
                    += (cont[i][j]
                        - (double)1)
                       / (double)2;
 
                cont[i + 1][j + 1]
                    += (cont[i][j]
                        - (double)1)
                       / (double)2;
            }
        }
    }
    cout << count;
}
 
// driver code
int main()
{
    int n = 3;
    double x = 5;
    num_of_containers(n, x);
    return 0;
}

Java

// Java program for the above problem
import java.util.*;
 
class GFG{
     
// Matrix of containers
static double cont[][] = new double[1000][1000];
 
// Function to find the number
// of containers that will be
// filled in X seconds
static void num_of_containers(int n, double x)
{
    int count = 0;
 
    // Container on top level
    cont[1][1] = x;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= i; j++)
        {
 
            // If container gets filled
            if (cont[i][j] >= (double)1)
            {
                count++;
 
                // Dividing the liquid
                // equally in two halves
                cont[i + 1][j] += (cont[i][j] -
                                   (double)1) /
                                   (double)2;
 
                cont[i + 1][j + 1] += (cont[i][j] -
                                       (double)1) /
                                       (double)2;
            }
        }
    }
    System.out.print(count);
}
 
// Driver code
public static void main(String[] args)
{
    int n = 3;
    double x = 5;
     
    num_of_containers(n, x);
}
}
 
// This code is contributed by jrishabh99

Python3

# Python3 program for the above problem
 
# Matrix of containers
cont = [[ 0 for i in range(1000)]
            for j in range(1000)]
             
# Function to find the number          
# of containers that will be          
# filled in X seconds          
def num_of_containers(n, x):         
     
    count = 0    
     
    # Container on top level          
    cont[1][1] = x    
     
    for i in range(1, n + 1):
        for j in range(1, i + 1):
             
             # If container gets filled          
             if (cont[i][j] >= 1):         
                count += 1         
                 
                # Dividing the liquid          
                # equally in two halves          
                cont[i + 1][j] += (cont[i][j] - 1) / 2         
                cont[i + 1][j + 1] += (cont[i][j] - 1) / 2
                 
    print(count)
     
# Driver code          
n = 3         
x = 5   
 
num_of_containers(n, x)
 
# This code is contributed by yatinagg

C#

// C# program for the above problem
using System;
 
class GFG{
     
// Matrix of containers
static double [,]cont = new double[1000, 1000];
 
// Function to find the number
// of containers that will be
// filled in X seconds
static void num_of_containers(int n, double x)
{
    int count = 0;
 
    // Container on top level
    cont[1, 1] = x;
    for(int i = 1; i <= n; i++)
    {
        for(int j = 1; j <= i; j++)
        {
             
            // If container gets filled
            if (cont[i, j] >= (double)1)
            {
                count++;
 
                // Dividing the liquid
                // equally in two halves
                cont[i + 1, j] += (cont[i, j] -
                                  (double)1) /
                                  (double)2;
 
                cont[i + 1, j + 1] += (cont[i, j] -
                                      (double)1) /
                                      (double)2;
            }
        }
    }
    Console.Write(count);
}
 
// Driver code
public static void Main(String[] args)
{
    int n = 3;
    double x = 5;
     
    num_of_containers(n, x);
}
}
 
// This code is contributed by Princi Singh

Javascript

输出

时间复杂度： O(N ² )
辅助空间复杂度： O(N ² )

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