POJ 1053 Integer Inquiry (大数加法,还是Java大法好)

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.) 

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. 

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

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

370370367037037036703703703670

East Central North America 1996

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		while(sc.hasNext()){
			BigInteger sum = new BigInteger("0",10);
			BigInteger x = sc.nextBigInteger();
			if(x.equals(new BigInteger("0"))){//注意,与'0'判断时要new对象!!!
				System.out.println(sum);
				break;
			}
			sum = sum.add(x);
		}

	}

}

AC代码2;

#include <stdio.h>
#include <string.h>
char a[1000];
int sum[1000],data[1000];
int length_sum;
int main()
{
    int i,j,length_a,jin,temp;
    memset(sum,0,sizeof(sum));
    while(1)
    {
        scanf("%s",a);
        length_a=strlen(a);
        if (a[0]=='0'&&length_a==1)
        {
            length_sum=999;
            while(sum[length_sum]==0)
            {
                length_sum--;
            }
            for (i=length_sum; i>=0; i--)
            {
                printf("%d",sum[i]);
            }
            break;
        }
        else
        {
            memset(data,0,sizeof(data));
            for (j=0,i=length_a-1; i>=0; i--)
            {
                data[j++]=a[i]-'0';
            }
            for(jin=0,i=0; i<1000; i++)
            {
                temp=sum[i]+data[i]+jin;
                sum[i]=temp%10;
                jin=temp/10;
            }
        }
    }
    return 0;
}