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#题目是给一个方法 foo(),能以各 50%的概率随机返回 0 或者 1。 | |
#问题是基于 foo()方法,实现一个 boo()方法,该方法能以各 10%的概率返回[0-9]中的一个数字。 | |
for i in range(10): | |
print("{:0>32b}".format(i), i) | |
import random | |
def foo(): | |
return random.randint(0,1) | |
# 0000 0 | |
# 0001 1 | |
# 0010 2 | |
# 0011 3 | |
# 0100 4 | |
# 0101 5 | |
# 0110 6 | |
# 0111 7 | |
# 1000 8 | |
# 1001 9 | |
def bar(): | |
s = [] | |
for _ in range(4): | |
s.append(foo()) | |
if s[0] and s[1]==0 and s[2]==0: | |
return 9 if s[3] else 8 | |
if s[0]==0 and s[1]: | |
if s[2]: | |
return 7 if s[3] else 6 | |
else: | |
return 5 if s[3] else 4 | |
if s[0]==0 and s[1]==0 and s[2]: | |
return 3 if s[3] else 2 | |
elif s[0]==0 and s[1]==0 and s[2] ==0: | |
return 1 if s[3] else 0 | |
return bar() | |
# test | |
count = {} | |
for i in range(10): | |
count[i] = 0 | |
for i in range(100000): | |
t = bar() | |
count[t] += 1 | |
for i in range(10): | |
print(i, count[i]) |
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