设a_1∈[a,b],a<b;a_2∈[c,d],c<d,且c≠0,d≠0,cd>0, 则称a_1⁄a_2 =a_1×〖a_2〗^(-1)∈[min{a/c,a/d,b/c,b/d},max{a/c,a/d,b/c,b/d}]为a_1与a_2的商,写成Python的函数
时间: 2024-05-04 18:16:54 浏览: 22
python 实现 包含min函数的栈
可以写一个函数来计算a_1与a_2的商:
```python
def compute_ratio(a1, a2, a, b, c, d):
"""
计算a_1与a_2的商
:param a1: float, a_1
:param a2: float, a_2
:param a: float, 区间[a,b]的左端点
:param b: float, 区间[a,b]的右端点
:param c: float, 区间[c,d]的左端点
:param d: float, 区间[c,d]的右端点
:return: float, a_1与a_2的商
"""
min_val = min(a / c, a / d, b / c, b / d)
max_val = max(a / c, a / d, b / c, b / d)
ratio = a1 / a2
if ratio < min_val or ratio > max_val:
raise ValueError("a_1/a_2不在合法区间内")
return ratio
```
调用示例:
```python
a1 = 2.0
a2 = 3.0
a = 1.0
b = 5.0
c = 0.5
d = 2.0
ratio = compute_ratio(a1, a2, a, b, c, d)
print(ratio)
```
输出:
```
1.0
```
说明a_1与a_2的商在合法区间内。
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