zoukankan      html  css  js  c++  java
  • hud1047 java大数忽略空格

    Integer Inquiry

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 10101    Accepted Submission(s): 2617


    Problem 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 only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.) 
     

    Input
    The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits (no VeryLongInteger will be negative). 

    The final input line will contain a single zero on a line by itself.
     

    Output
    Your program should output the sum of the VeryLongIntegers given in the input. 


    This problem contains multiple test cases!

    The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

    The output format consists of N output blocks. There is a blank line between output blocks.
     

    Sample Input
    1 123456789012345678901234567890 123456789012345678901234567890 123456789012345678901234567890 0
     

    Sample Output
    370370367037037036703703703670
    package com.acm.vivi;
    
    import java.math.BigInteger;
    import java.util.Scanner;
    
    public class Main {
    	public static void main(String []args)throws Exception{
    		Scanner cin = new Scanner(System.in);
    		int t = cin.nextInt();
    		while(t--!=0){
    		BigInteger sum = BigInteger.ZERO;
    		while(true){
    			String s = cin.next();
    			if(s.equals("0"))
    				break;
    			sum=sum.add(new BigInteger(s));
    		}
    		System.out.println(sum);
    		if(t>0)
    			System.out.println();
    		}
    		cin.close();
    	}
    }
    

    //java输入就不用考虑空行了
  • 相关阅读:
    Python+Selenium自动化篇-3-设置浏览器大小、刷新页面、前进和后退
    Python+Selenium自动化篇-2-定位页面元素的八种方法
    Python+Selenium自动化篇-1-安装模块和浏览器驱动操作方法
    02_Python_数据容器(列表、元祖、集合、字典)
    01_Python_基础知识(变量、字符串、条件控制)
    orm单表操作
    模版导入与继承
    django模板层
    orm视图函数层
    orm路由层
  • 原文地址:https://www.cnblogs.com/unclejelly/p/4082109.html
Copyright © 2011-2022 走看看