1010: 折半查找的实现（python）

count=0
nums=[]
n=input()
s=input()
target=int(input())
for i in s.split(" "):
    nums.append(int(i))
flag=0
left=0
right=len(nums)-1
while(left<=right):
    mid=left+(right-left)//2
    if nums[mid]>target:
        right=mid-1
        count=count+1
    elif nums[mid]<target:
        left=mid+1
        count = count + 1
    else:
        flag=1
        print(mid)
        break
if flag==0:
    print(-1)
print(count,end="")#真傻逼，没有这个会格式报错