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onder This Challenge:
When you run the following pseudo code on a sequence of six bits, in exactly 14 cases you get a non-zero value. How many such cases would you get for all sequences of 42 bits?
s=2 r=0 for b in v: if (s modulo 7) = b: r = r+2*b-1 s = 6+s*(5+s*s*s*(2+s*(3+s))) + b*(5+s*(5+s*(6+s*s*(3+s*6)))) return r
Bonus question:If we would have asked a similar question for sequences of 14 bits, what would be the connection of the result to IBM?
Hint:We asked a similar bonus question in the past, and the number of solvers was then a multiplication of two prime numbers.


IP属地:浙江来自Android客户端17楼2017-02-03 21:36
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    ”这一挑战:
    当你在六位序列上运行下面的伪代码时,在恰好的14种情况下,你得到一个非零值。有多少这样的情况下,你会得到所有序列的42位?
    S = 2
    R = 0
    在乙中:
    如果(S模7)= B:
    R = r + 2×B-1
    s = 6 + s *(5 + s s * * * *(2 + s *(3 + s))+ B *(5 + s *(5 + s *(6 + s * * * * * * * * 6))
    返回R
    另外一个问题:如果我们有14位序列问过类似的问题,会有什么结果IBM连接吗?
    提示:我们过去曾问过类似的奖金问题,而求解器的数目则是两个素数的乘法。


    IP属地:浙江来自Android客户端18楼2017-02-03 21:38
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      抓鸡啊… 应该是那类我不会做的题型


      IP属地:浙江来自Android客户端19楼2017-02-03 21:41
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        计算机文盲题目都看不懂…… 看懂了估计也解不出 数学谜题我基本做不出…


        IP属地:浙江来自Android客户端20楼2017-02-05 23:22
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