科大讯飞C++方向两道AC代码(python)
作者：ukulele201903082324789
链接：https://www.nowcoder.com/discuss/225495?type=post&order=time&pos=&page=1&subType=1
来源：牛客网

1.割木棍：
n = int(input())
ret = 0

if n == 2:
print(1)
elif n == 3:
print(2)
else:
for i in range(0,100):
for j in range(0,100):
if&nbsp;i*2&nbsp;+&nbsp;j*3&nbsp;==&nbsp;n&nbsp;and&nbsp;ret&nbsp;<&nbsp;(2**i)*(3**j):
ret&nbsp;=&nbsp;(2**i)*(3**j)
print(ret)

2.二分查找
n&nbsp;=&nbsp;int(input())
a&nbsp;=&nbsp;list(map(int,input().split()))
b&nbsp;=&nbsp;int(input())
L&nbsp;=&nbsp;0
R&nbsp;=&nbsp;n-1
ret&nbsp;=&nbsp;-1
while&nbsp;L&nbsp;<=&nbsp;R:
mid&nbsp;=&nbsp;(L&nbsp;+&nbsp;R)&nbsp;//2
if&nbsp;a[mid]&nbsp;>=&nbsp;b:
ret&nbsp;=&nbsp;mid
R&nbsp;=&nbsp;mid&nbsp;-&nbsp;1
else:
L&nbsp;=&nbsp;mid&nbsp;+&nbsp;1
if&nbsp;ret&nbsp;==&nbsp;-1&nbsp;or&nbsp;a[ret]&nbsp;!=&nbsp;b:
print(-1)
else:
print(ret)