原题地址 竟然64位都要爆,这是要大整数乘法的节奏吗?我才不要写大整数乘法呢,用Ruby干掉 代码: # Enter your code here. Read input from STDIN. Print output to STDOUT num = [0, 0] num[0], num[1], n = readline.chomp.split(' ').map {|e| e.to_i} (n - 2).times do tmp = num[1] ** 2 + num[0] num[0] =…

题目来源:Fibonacci Modified We define a modified Fibonacci sequence using the following definition: Given terms  and  where , term  is computed using the following relation:   For example, if term  and , term , term , term , and so on. Given three intege…

Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9394    Accepted Submission(s): 3065 Problem Description A Fibonacci sequence is calculated by adding the previous two members the…

题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1250 hdu1250: Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9442    Accepted Submission(s): 3096 Problem Description A Fibonacci…

大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11520 Accepted Submission(s): 3911 Problem Description Fibonacci数列,定义如下:f(1)=f(2)=1f(n)=f(n-1)+f(n-2) n>=3.计算第n项Fibonacci数值.   Input 输入第一行为一个整数N…

传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1250 Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 12952    Accepted Submission(s): 4331 Problem Description A Fibonacci sequence…

/*******对读者说(哈哈如果有人看的话23333)哈哈大杰是华农的19级软件工程新手,才疏学浅但是秉着校科联的那句“主动才会有故事”还是大胆的做了一下建一个卑微博客的尝试,想法自己之后学到东西都记录一下自己学的同时或许(我说或许啊哈哈)能帮到博友,如果有啥错误的话还请各位大佬在下面留言怼我,指出我的错误所在,我一定更改哈哈,一般记录的都是我对一个知识点或者是一个算法专题的笔记和一些在博客园里面看到写的好的大佬的一些借鉴文章大部分都是在codeblocks里面写好了然后复制过来的,所以就有很…

The well-known Fibonacci sequence is defined as following: F(0) = F(1) = 1 F(n) = F(n − 1) + F(n − 2) ∀n ≥ 2 Here we regard n as the index of the Fibonacci number F(n). This sequence has been studied since the publication of Fibonacci’s book Liber Ab…

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int a[7500][670];void dashu(){ memset(a,0,sizeof(a)); a[1][1]=1, a[2][1]=1, a[3][1]=1, a[4][1]=1; int k=0; for(int i=5;i<7500;i++) for(int j=1;j<=670;j++)…

思路 用PYTHON 或 JAVA 干掉 AC代码 a, b, n = map(int, input().split()) for i in range (2, n, 1) : temp = b b = b ** 2 + a a = temp print(b)…