//  继续大数,哎.. 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 tas…

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 >…

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…

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…

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…

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 java做大数的题,真的是神器,来一道,秒一道~~~ import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @xqq */ public BigInteger an(int n) { BigInteger e; BigInteger a = BigInteger.valueOf(1); BigInteger b = BigInteger.valueOf(1); BigInteg…

pid=1250">点击此处就可以传送hdu 1250 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…

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 输入第一行为一个整数N,接下来N行为整数Pi(1<=Pi<=1000). Output 输出为N行,每行为对应的f(Pi). Sample Input 5 1 2 3 4 5 Sample Output 1 1 2 3 5 /*************** 继续大数. 用 hdu 1047 写…

题意:对给定前缀(长度不超过40),找到一个最小的n,使得Fibonacci(n)前缀与给定前缀相同,如果在[0,99999]内找不到解,输出-1. 思路:用高精度加法计算斐波那契数列,因为给定前缀长度不超过40,所以高精度计算时每次只需保留最高60位,每次将得到的值插入到字典树中,使得树上每个节点只保留最小的n值.查询输出字典树结点的值. #include<cstdio> #include<cstring> #include<iostream> #define MAX…