Codeforces #307(Div. 2) Problem D. GukiZ and Binary Operations

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这个题刚开始是看错题了。。一点思路都没有

到后来把题目看对了以后，有一点思路，但还是不会做。。无奈只能看官方题解

官方题解是这样的：

First convert number k into binary number system. If some bit of k is 0 than the result of or opertion applied for every adjacent pair of those bits in array a must be 0, that is no two adjacent those bits in array a are 1. We should count how many times this is fulfilled. If the values were smaller we could count it with simply dp_i = dp_i - 1 + dp_i - 2, where dp_i is equal to number of ways to make array od i bits where no two are adjacent ones. With first values dp₁ = 2 and dp₂ = 3, we can see that this is ordinary Fibonacci number. We can calculate Fibonacci numbers up to 10¹⁸ easily by fast matrix multiplication. If some bit at k is 1 than number of ways is 2ⁿ — \t{(number of ways bit is 0)}, which is also easy to calculate. We must be cearful for cases when 2^l smaller than k (solution is 0 then) and when l = 63 or l = 64. Total complexity is .

看了以后大概懂了他的思路。将每一二进制位看成一步，则根据乘法原理，总方案数=每一位的方案数的积。

如果k的某一位二进制值为零，则发现每个相邻元素不能均为1.然后dp即可，是斐波那契数列，矩阵乘法O(log n) 无压力

如果k的某一位二进制值为1，则必然存在两个相邻元素的值为1，然后发现f[n]=2^(n-1)+f[n-1] 答案为2*f[n-1]，然而这个式子让人很不爽。考虑到这是计数类问题，从反面考虑，则方案为2^n-dp[n]( dp[n]是指某一二进制位为零时其取值 ），然后就爽了。。

Codeforces #307(Div. 2) Problem D. GukiZ and Binary Operations

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原文地址：http://www.cnblogs.com/tsingyawn/p/4576322.html