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…
Problem Description A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1. F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4) Your task is to take…
题目:http://acm.hdu.edu.cn/showproblem.php?pid=1250 思路:本题的Fibonacci数列是扩展的四阶的Fibonacci数列,用递推关系式求解即可. 题目提示,所求的Fibonacci数最多2500位,所以不能用已有的数据类型表示,可以采用数组存储数字,模拟加法. 由于没有给F(n)的n最大为多少,所以需要估计一下.经过反复测试,F(7000)约为2000位左右,如果设置数组长度为2500+1,则计算得到F(7500)的结果沾满了2500+1位(),…
Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14776    Accepted Submission(s): 4923   Problem Description A Fibonacci sequence is calculated by adding the previous two members…
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…
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…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1250 Hat's Fibonacci Description A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.$F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n >…
Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 5800    Accepted Submission(s): 1926 Problem Description A Fibonacci sequence is calculated by adding the previous two members the…
大菲波数 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16125    Accepted Submission(s): 5355 Problem Description Fibonacci数列,定义如下: f(1)=f(2)=1 f(n)=f(n-1)+f(n-2) n>=3. 计算第n项Fibonacci数值.   Input 输…
UVa 495 求第n位斐波那契数列,n<=5000. 还是大数问题,这次是大数加法.仿照UVa 623的解法来做.623位数可以一位一位的增,但是这个需要预先给够位数,要是按六位存一个数组元素里面的话,300位足够了,粗算一下n=5000大概有1044位,len=300*6足够存下了.就是不知道最后输出答案时我让pos初始为len为什么会WA,这个初始化对结果有影响? #include<iostream> #include<cstdio> #define mod 10000…