题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1047 题目描述: 题意有点绕,但是仔细的读了后就发现是处理大数相加的问题.注意:输入数据有多组,每组输入要用一个t来控制输入的模块,每个模块的输入以0来结束.换句话说,有多少模块,输出就有多少个VeryLongInteger. 代码实现: import java.math.BigInteger; import java.util.Scanner; public class Hdu1047 { pub…

import java.math.BigInteger; import java.util.Scanner; public class Bignum{    public static void main(String[] args) {        Scanner s=new Scanner(System.in);        BigInteger a=new BigInteger("0");        BigInteger b=new BigInteger("0&…

大菲波数 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 17352 Accepted Submission(s): 5778 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…

package Main; //import java.io.InputStream; import java.math.BigDecimal; import java.util.Scanner; public class Main { public static void main(String[] args) { int _case; BigDecimal a,b; a=new BigDecimal("0"); b=new BigDecimal("0"); Sc…

题目大意:输入n组数据,每组数据中又有若干长度不大于100的整数,以0结束每组数据的输入,求每组中数据之和.每两组数据输入之间有一行空格,输出也是如此. Integer Inquiry Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 17770 Accepted Submission(s): 4628 Problem Description O…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1250 题目描述: 题目大意是:已知一个Hat's Fibonacci序列,该序列满足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); 此题用java的BigInteger处理很方便,但是因为我在处理F(n>4) = F(n - 1) + F(n-2) + F(n-3) + F(n…

题目 我要开始练习一些java的简单编程了^v^ import java.io.*; import java.util.*; import java.math.*; public class Main { /** * @param args */ public static void main(String[] args) throws Exception { // 定义并打开输入文件 Scanner cin = new Scanner(System.in); BigInteger a, sum…

一.Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers. ``This supercomputer is great,'' remarked Chip. ``I…

大数相加: package algorithm; //使用BigInteger类验证 import java.math.BigInteger; public class BigAdd { public static String bigNumberAdd(String f, String s) { // 翻转两个字符串,并转换成数组 char[] a = new StringBuffer(f).reverse().toString().toCharArray(); char[] b = new…

一.Description The creators of a new programming language D++ have found out that whatever limit for SuperLongInt type they make, sometimes programmers need to operate even larger numbers. A limit of 1000 digits is so small... You have to find the sum…