Integer Inquiry_hdu_1047(大数).java
发布时间:2020-12-14 04:02:23 所属栏目:大数据 来源:网络整理
导读:Integer Inquiry Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9376????Accepted Submission(s): 2406 Problem Description One of the first users of BIT's new supercomputer was Chip Di
Integer InquiryTime Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9376????Accepted Submission(s): 2406
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.)?
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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.
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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.
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Sample Input
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Sample Output
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Source
East Central North America 1996
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import java.math.BigInteger; import java.util.Scanner; public class Main {//没咋看懂是怎么输入的 public static void main(String[] args) { Scanner input=new Scanner(System.in); while(input.hasNext()){ int n=input.nextInt(); while(n-->0){ BigInteger sum=BigInteger.valueOf(0); BigInteger a; while(!(a=input.nextBigInteger()).equals(BigInteger.valueOf(0))){ sum=sum.add(a); } System.out.println(sum); if(n>0) System.out.println(); } } } } (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |