https://blog.paulhankin.net/fibonacci/ This code, somewhat surprisingly, generates Fibonacci numbers. def fib(n): return (4 << n*(3+n)) // ((4 << 2*n) - (2 << n) - 1) & ((2 << n) - 1) In this blog post, I'll explain where it co…

Fibonacci Numbers Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 81 Accepted Submission(s): 46   Problem Description The Fibonacci sequence is the sequence of numbers such that every element is e…

这是个开心的题目,因为既可以自己翻译,代码又好写ヾ(๑╹◡╹)ﾉ" The i’th Fibonacci number f(i) is recursively defined in the following way: • f(0) = 0 and f(1) = 1 • f(i + 2) = f(i + 1) + f(i) for every i ≥ 0 Your task is to compute some values of this sequence. Input Input begins…

The Fibonacci numbers are the numbers in the following integer sequence. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, …….. In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation Fn = Fn-1 + Fn-2with seed va…

In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation F1 = 1; F2 = 1; Fn = Fn - 1 + Fn - 2 (n > 2). DZY loves Fibonacci numbers very much. Today DZY gives you an array consisting of n integers: a1, a2, ...,…

C. DZY Loves Fibonacci Numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation F1 = 1; F2 …

C. DZY Loves Fibonacci Numbers time limit per test 4 seconds memory limit per test 256 megabytes input standard input output standard output In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation F1 = 1; F2 …

參考:http://www.cnblogs.com/chanme/p/3843859.html 然后我看到在别人的AC的方法里还有这么一种神方法,他预先设定了一个阈值K,当当前的更新操作数j<K的时候,它就用一个类似于树状数组段更的方法,用一个 d数组去存内容,譬如它要在区间 [3,6]上加一段fibonacci 原来: id 0 1 2 3 4 5 6 7 8 9 10 d  0 0 0 0 0 0 0 0 0 0 0 更新: id 0 1 2 3 4 5 6  7  8  9 10 d  0…

HDU 3117 Fibonacci Numbers(斐波那契前后四位,打表+取对+矩阵高速幂) ACM 题目地址:HDU 3117 Fibonacci Numbers 题意:  求第n个斐波那契数的前四位和后四位.  不足8位直接输出. 分析:  前四位有另外一题HDU 1568,用取对的方法来做的.  后四位能够用矩阵高速幂,MOD设成10000即可了. 代码: /* * Author: illuz <iilluzen[at]gmail.com> * Blog: http://blog.c…

Portal Description 给出一个\(n(n\leq3\times10^5)\)个数的序列,进行\(m(m\leq3\times10^5)\)次操作,操作有两种: 给区间\([L,R]\)加上一个斐波那契数列,即\(\{a_L,a_{L+1},...,a_R\} \rightarrow \{a_L+F_1,a_{L+1}+F_2,...,a_R+F_{R-L+1}\}\) 询问区间\([L,R]\)的和,对\(10^9+9\)取模. 斐波那契数列:\(F_1=1,F_2=2\)且满足…