HDU 1250 Hat's Fibonacci(高精度)】的更多相关文章

//  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…
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…
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…