hdu1047(Java)大数相加

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

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

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import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
public class Main1047 {
public static void main(String[] args) {
Scanner cin=new Scanner(System.in);
while(cin.hasNext()){
int n=cin.nextInt();
Object obj=BigInteger.ZERO;
for(int i=0;i<n;i++){
BigInteger sum=BigInteger.ZERO;
while(true){
BigInteger b=cin.nextBigInteger();
if(b.equals(obj)){//这里也可以用commparTo(BigInteger.ZERO)==0进行判断；
System.out.println(sum);
if(i!=n-1)
System.out.println();
break;
}
sum=sum.add(b);
}
}
}
}
}