编写函数,输出如图杨辉三角,并运行验证(使用python)
时间: 2024-05-05 08:21:29 浏览: 39
可以使用以下代码来输出杨辉三角:
```python
def pascal_triangle(n):
triangle = []
for i in range(n):
row = [1]
if i > 0:
prev_row = triangle[-1]
for j in range(len(prev_row)-1):
row.append(prev_row[j] + prev_row[j+1])
row.append(1)
triangle.append(row)
return triangle
# 输出杨辉三角
n = 10 # 三角形的行数
triangle = pascal_triangle(n)
for row in triangle:
print(row)
```
运行以上代码可以得到输出结果:
```
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
```
可以发现,输出结果与杨辉三角的规律是相符的,验证了代码的正确性。
相关问题
编写函数,输出如图杨辉三角。并运行验证
非常感谢您的提问。以下是一个Python函数,可用于打印杨辉三角:
```
def generate_pascal_triangle(n):
# 初始化杨辉三角列表
pascal_triangle = [[1]]
# 生成杨辉三角
for i in range(1, n):
row = [1]
for j in range(1, i):
row.append(pascal_triangle[i-1][j-1] + pascal_triangle[i-1][j])
row.append(1)
pascal_triangle.append(row)
# 输出杨辉三角
for i in range(n):
# 打印空格
for j in range(n-i-1):
print(" ", end="")
# 打印数字
for j in range(i+1):
print("{:2d} ".format(pascal_triangle[i][j]), end="")
print()
```
在这里,我们需要为函数提供一个整数 n 作为杨辉三角的行数。函数使用双重循环来生成杨辉三角并打印出来。在输出时,我们使用 format 格式化字符串来使数字保持对齐。
要测试这个函数,您可以将其复制并粘贴到Python解释器中,然后调用它并提供所需的参数。例如,要打印一个四行的杨辉三角,您可以这样做:
```
generate_pascal_triangle(4)
```
这将打印出以下内容:
```
1
1 1
1 2 1
1 3 3 1
```
希望这可以帮助您解决您的问题!
编写函数,输出六行杨辉三角,并运行验证(使用python)
可以使用以下代码来打印六行杨辉三角:
```python
def generate_pascal_triangle(n):
triangle = []
for i in range(n):
row = [1] * (i+1) # 初始化每一行的第一个和最后一个元素都为1
if i > 1:
for j in range(1, i):
row[j] = triangle[i-1][j-1] + triangle[i-1][j]
triangle.append(row)
return triangle
def print_pascal_triangle(n):
triangle = generate_pascal_triangle(n)
for row in triangle:
print(row)
print_pascal_triangle(6)
```
输出结果:
```
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
```
可以看到,输出了六行杨辉三角。
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