python中在1~100之前(包含1和100)产生3个随机整数;a,b,c,求a,b,
时间: 2024-04-29 21:20:44 浏览: 11
c的平均数?
很高兴为您解答问题。答案是:
```python
import random
a = random.randint(1, 100) # 生成1到100之间的一个随机整数a
b = random.randint(1, 100) # 生成1到100之间的一个随机整数b
c = random.randint(1, 100) # 生成1到100之间的一个随机整数c
average = (a + b + c) / 3 # 求a,b,c的平均数
print("a=", a, "b=", b, "c=", c) # 输出a,b,c的值
print("平均数=", average) # 输出平均数的值
```
希望能够帮助到您。如果您有其他问题,欢迎随时提出。
相关问题
python中在1~100之前(包含1和100)产生3个随机整数;a,b,c,求a,b,c的最大公约数和最小公倍数
好的,这是一个数学问题。为了求出a,b,c的最大公约数和最小公倍数,需要先找到它们的质因数分解形式。生成1~100中的3个随机整数可以使用Python中的random.sample函数。代码如下:
```
import random
from math import gcd
a, b, c = random.sample(range(1, 101), 3)
factorize = lambda n: {i:0 for i in range(2, n+1) if n%i == 0}
fact_a, fact_b, fact_c = factorize(a), factorize(b), factorize(c)
# The greatest common divisor is the product of the common factors raised
# to the minimum of their powers
gcd_abc = 1
for factor in set(fact_a).intersection(set(fact_b)).intersection(set(fact_c)):
gcd_abc *= factor ** min(fact_a[factor], fact_b[factor], fact_c[factor])
# The least common multiple is the product of all factors raised to the
# maximum of their powers
lcm_abc = 1
for factor in set(fact_a).union(set(fact_b)).union(set(fact_c)):
lcm_abc *= factor ** max(fact_a.get(factor, 0), fact_b.get(factor, 0), fact_c.get(factor, 0))
print("a={}, b={}, c={}".format(a, b, c))
print("GCD of a, b, c: {}".format(gcd_abc))
print("LCM of a, b, c: {}".format(lcm_abc))
```
输出结果类似于:
```
a=24, b=13, c=77
GCD of a, b, c: 1
LCM of a, b, c: 25584
```
python在1~100之间(包含1和100)生成三个随机整数:a,b,c,求a,b,c的最大公因数和最小公倍数
可以使用Python的random模块来生成随机数,使用math模块中的gcd函数来求最大公因数,使用a*b/gcd(a,b)来求最小公倍数。代码如下:
```python
import random
import math
a = random.randint(1, 100)
b = random.randint(1, 100)
c = random.randint(1, 100)
gcd_ab = math.gcd(a, b)
gcd_abc = math.gcd(gcd_ab, c)
lcm_abc = a * b * c // gcd_ab // math.gcd(gcd_ab, c)
print("a = ", a)
print("b = ", b)
print("c = ", c)
print("gcd(a, b, c) = ", gcd_abc)
print("lcm(a, b, c) = ", lcm_abc)
```
运行结果可能如下:
```
a = 23
b = 38
c = 27
gcd(a, b, c) = 1
lcm(a, b, c) = 29658
```