题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(2^n)$

暴力枚举找规律

时间复杂度

参考文献

python 代码

for _ in range(2,10):
    l = pow(2,_)
    r = pow(2,_ + 1)
    ans = 0
    for i in range(0,r):
        f = [int(j) for j in (bin(i)[2:]).zfill(_ + 1)]
        cnt = 0
        n = len(f)
        flag = 1
        print(f)
        for x in range(n):
            hm = f[(x + 1) % n]
            back = f[(x + 2) % n]
            #如果后面一个为真一个为假
            if f[x] == 1:
                if not ((hm == 1 and back == 0) or (hm == 0 and back == 1)):
                    flag = 0
                    break
            else:
                if ((hm == 1 and back == 0) or (hm == 0 and back == 1)):
                    flag = 0
                    break
                cnt += 1
        if flag:
            ans += cnt
    print(ans)

算法2

(暴力枚举) $O(1)$

打表可得n为3的倍数时答案为n * 2否则为n

时间复杂度

参考文献

C++ 代码

for _ in range(int(input())):
    n = int(input())
    print(n * 2 if n % 3 == 0 else n)