UVA11040 Add bricks in the wall【数学】

This in not “another brick in the wall”, it’s just a matter of adding numbers. Suppose you have a wall with the shape of a triangle, like the one shown below. The wall has 9 rows and row i has exactly i bricks, considering that top row is the first one and bottom row is the ninth. Some bricks are labeled with a number and other ones are blank. Notice that labeled bricks appear only on odd rows and they occupy odd positions within the row. The problem you must solve is finding a suitable number for each blank brick taking into account one simple rule: the number of a brick is obtained by adding the numbers of the two bricks below it. Obviously, this rule does not apply to the ninth row. Numbers are supposed to be integers.

Input

The first line of the input contains an integer N, indicating the number of test cases. This line is followed by the lines corresponding to the test cases. Each test case is described in five lines. These five lines correspond to odd rows of the wall, from top to bottom, as described above. Line i contains the numbers corresponding to odd bricks on row i of the wall (that is, non blank bricks), enumerated from left to right and separated with a single space. It is supposed that each test case is correct, that is, there exists a solution to the problem that the case describes.

Output

For each test case, the output should consist of nine lines describing the numbers of all bricks of the wall. So, line i should contain the numbers corresponding to the i bricks on row i of the wall, enumerated from left to right and separated by a single space.

Note: Here we have an example with two test cases. The first one corresponds to the wall depicted above.

Sample Input

255

54 67

10 18 13

3 3 5 2

2 1 2 1 1

256

64 64

16 16 16

4 4 4 4

1 1 1 1 1

Sample Output

255

121 134

54 67 67

23 31 36 31

10 13 18 18 13

5 5 8 10 8 5

3 2 3 5 5 3 2

2 1 1 2 3 2 1 1

2 0 1 0 2 1 1 0 1

256

128 128

64 64 64

32 32 32 32

16 16 16 16 16

8 8 8 8 8 8

4 4 4 4 4 4 4

2 2 2 2 2 2 2 2

1 1 1 1 1 1 1 1 1

问题链接：UVA11040 Add bricks in the wall

问题简述：（略）

问题分析：

这个问题首先是数据表示问题。三角形是不方便的，只好用二维数组表示，并且左对齐，将近一半就浪费了。这也是没有办法的。

然后就是找规律，观察以下的三角形数据结构关系：

[c]

[a+x] [x+b]

[a] [x] [b]

其中a,b和c是已知的，x是未知的，那么c=(a+x)+(x+b)，得x=(c-a-b)/2。

用上述的x公式，就可以计算下标为偶数（第0行不用算）的行的未知项。然后再计算奇数行的各项。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* UVA11040 Add bricks in the wall */#include <bits/stdc++.h>//#define DEBUGusing namespace std;const int N = 9;
int a[N][N];void print()
{for(int i=0; i<N; i++) {for(int j=0; j<=i; j++) {if(j)printf(" ");printf("%d", a[i][j]);}printf("\n");}
}int main()
{int t;#ifdef DEBUGmemset(a, 0, sizeof(a));
#endifscanf("%d", &t);while(t--) {// 读入数据for(int i=0; i<N; i+=2)for(int j=0; j<=i; j+=2)scanf("%d", &a[i][j]);#ifdef DEBUGprint();
#endif// 计算偶数行for(int i=N-1; i>0; i-=2)for(int j=1; j<i; j+=2)a[i][j] = (a[i - 2][j - 1] - a[i][j - 1] - a[i][j + 1]) / 2;#ifdef DEBUGprint();
#endif// 计算奇数行for(int i=1; i<N; i+=2)for(int j=0; j<=i; j++)a[i][j] = a[i+1][j] + a[i+1][j+1];print();}return 0;
}

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UVA11040 Add bricks in the wall【数学】

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