solution.md

# 承压计算

X星球的高科技实验室中整齐地堆放着某批珍贵金属原料。

每块金属原料的外形、尺寸完全一致，但重量不同。  
金属材料被严格地堆放成金字塔形。  
                        

```

                             7 
                            5 8 
                           7 8 8 
                          9 2 7 2 
                         8 1 4 9 1 
                        8 1 8 8 4 1 
                       7 9 6 1 4 5 4 
                      5 6 5 5 6 9 5 6 
                     5 5 4 7 9 3 5 5 1 
                    7 5 7 9 7 4 7 3 3 1 
                   4 6 4 5 5 8 8 3 2 4 3 
                  1 1 3 3 1 6 6 5 5 4 4 2 
                 9 9 9 2 1 9 1 9 2 9 5 7 9 
                4 3 3 7 7 9 3 6 1 3 8 8 3 7 
               3 6 8 1 5 3 9 5 8 3 8 1 8 3 3 
              8 3 2 3 3 5 5 8 5 4 2 8 6 7 6 9 
             8 1 8 1 8 4 6 2 2 1 7 9 4 2 3 3 4 
            2 8 4 2 2 9 9 2 8 3 4 9 6 3 9 4 6 9 
           7 9 7 4 9 7 6 6 2 8 9 4 1 8 1 7 2 1 6 
          9 2 8 6 4 2 7 9 5 4 1 2 5 1 7 3 9 8 3 3 
         5 2 1 6 7 9 3 2 8 9 5 5 6 6 6 2 1 8 7 9 9 
        6 7 1 8 8 7 5 3 6 5 4 7 3 4 6 7 8 1 3 2 7 4 
       2 2 6 3 5 3 4 9 2 4 5 7 6 6 3 2 7 2 4 8 5 5 4 
      7 4 4 5 8 3 3 8 1 8 6 3 2 1 6 2 6 4 6 3 8 2 9 6 
     1 2 4 1 3 3 5 3 4 9 6 3 8 6 5 9 1 5 3 2 6 8 8 5 3 
    2 2 7 9 3 3 2 8 6 9 8 4 4 9 5 8 2 6 3 4 8 4 9 3 8 8   
   7 7 7 9 7 5 2 7 9 2 5 1 9 2 6 5 3 9 3 5 7 3 5 4 2 8 9   
  7 7 6 6 8 7 5 5 8 2 4 7 7 4 7 2 6 9 2 1 8 2 9 8 5 7 3 6   
 5 9 4 5 5 7 5 5 6 3 5 3 9 5 8 9 5 4 1 2 6 1 4 3 5 3 2 4 1 
X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X 
```

其中的数字代表金属块的重量（计量单位较大）。  
最下一层的X代表30台极高精度的电子秤。  

假设每块原料的重量都十分精确地平均落在下方的两个金属块上，  
最后，所有的金属块的重量都严格精确地平分落在最底层的电子秤上。  
电子秤的计量单位很小，所以显示的数字很大。  

工作人员发现，其中读数最小的电子秤的示数为：```2086458231```  

请你推算出：读数最大的电子秤的示数为多少？  


## aop

### before

```cpp
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std;
typedef long long ll;
const ll maxn = 1e6 + 50;
double a[1050][1050];

```
### after

```cpp

```

## 答案

```cpp
int main()
{
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			scanf("%lf", &a[i][j]);
		}
	}
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			a[i + 1][j] += (0.5 * a[i][j]);
			a[i + 1][j + 1] += (0.5 * a[i][j]);
		}
	}
	double minn = 1000000000, maxx = -1;
	for (ll i = 1; i <= 30; i++)
	{
		maxx = max(maxx, a[30][i]);
		minn = min(minn, a[30][i]);
		printf("%lf\n", a[30][i]);
	}
	printf("%lf\n", maxx * 2086458231 / minn);
}
```
## 选项


### A

```cpp

int main()
{
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			scanf("%lf", &a[i][j]);
		}
	}
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			a[i + 1][j] += (0.5 * a[i][j]);
			a[i + 1][j + 1] += (0.5 * a[i][j]);
		}
	}
	double minn = 1000000000, maxx = -1;
	for (ll i = 1; i <= 29; i++)
	{
		maxx = max(maxx, a[30][i]);
		minn = min(minn, a[30][i]);
		printf("%lf\n", a[30][i]);
	}
	printf("%lf\n", maxx * 2086458231 / minn);
}
```

### B

```cpp
int main()
{
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			scanf("%lf", &a[i][j]);
		}
	}
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			a[i + 1][j] = (0.5 * a[i][j]);
			a[i + 1][j + 1] = (0.5 * a[i][j]);
		}
	}
	double minn = 1000000000, maxx = -1;
	for (ll i = 1; i <= 29; i++)
	{
		maxx = max(maxx, a[30][i]);
		minn = min(minn, a[30][i]);
		printf("%lf\n", a[30][i]);
	}
	printf("%lf\n", maxx * 2086458231 / minn);
}
```

### C

```cpp
int main()
{
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			scanf("%lf", &a[i][j]);
		}
	}
	for (ll i = 1; i <= 29; i++)
	{
		for (ll j = 1; j <= i; j++)
		{
			a[i + 1][j] += (0.5 * a[i][j]);
			a[i][j + 1] += (0.5 * a[i][j]);
		}
	}
	double minn = 1000000000, maxx = -1;
	for (ll i = 1; i <= 30; i++)
	{
		maxx = max(maxx, a[30][i]);
		minn = min(minn, a[30][i]);
		printf("%lf\n", a[30][i]);
	}
	printf("%lf\n", maxx * 2086458231 / minn);
}

```