题目 //f(1) = 1, f(2) = 1, f(n > 2) = f(n - 1) + f(n - 2) import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @xqq */ public BigInteger an(int n) { BigInteger c; BigInteger a = BigInteger.valueOf(1); BigInteger b = BigIn…

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…

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…

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 …

主题链接:CLICK HERE~ 有了Java求解大数变得如此简单,以后再也不用操心大数模板了.哦啦啦啦. import java.math.BigInteger; import java.math.BigDecimal; import java.util.Scanner; class Main{ public static void main(String args[]){ Scanner cin = new Scanner(System.in); while(cin.hasNext()){…

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…

题目:DZY Loves Fibonacci Numbers 题意比較简单,不解释了. 尽管官方的题解也是用线段树,但还利用了二次剩余. 可是我没有想到二次剩余,然后写了个感觉非常复杂度的线段树,还是mark一下吧. 我是这样考虑这个问题的,首先准备三个数组F,G,K,功能后面解释. 然后对它们有这样一个计算: F[0] = G[0] = 0; F[1] = 1; G[1] = 0; K[0] = 1; K[1] = 0; for(int i=2; i<N; i++){ F[i] = (F[i-…

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…

原题:http://acm.hdu.edu.cn/showproblem.php?pid=5686 当我们要求f[n]时,可以考虑为前n-1个1的情况有加了一个1. 此时有两种情况:当不适用第n个1进行合并时,就有f[n-1]个序列:当使用这个1进行合并时,就有f[n-2]个序列.所以f[n] = f[n-1]+f[n-2]. 因为这道题数会很大,所以可以用Java做大数运算. import java.math.BigInteger; import java.util.Scanner; publ…

java简单词法分析器 : http://files.cnblogs.com/files/hujunzheng/%E7%AE%80%E5%8D%95%E8%AF%8D%E6%B3%95%E5%88%86%E6%9E%90%E5%99%A8.zip…