Integer?Inquiry

Time?Limit:?2000/1000?MS?(Java/Others)????Memory?Limit:?65536/32768?K?(Java/Others)
Total?Submission(s):?13201????Accepted?Submission(s):?3322

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

1??1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678900

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Sample?Output

370370367037037036703703703670

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#include<stdio.h>
#include<string.h>
#define N 110



int main()
{
	int t,sum[N],rnum[N],len;
	char num[N];
	scanf("%d",&t);
	while(t--)
	{
		memset(num,'',sizeof(num));
		memset(sum,sizeof(sum));
		
		while(scanf("%s",num)!=EOF)
		{
			if(strcmp(num,"0")==0)
			{
				break;
			}
			
			memset(rnum,sizeof(rnum));
			len = strlen(num);

			for(int i=0;i<len;i++)
			{
				rnum[i] = num[len-1-i]-'0';
			}
			
			memset(num,sizeof(num));
			for(int i=0;i<N;i++)
			{
				sum[i] += rnum[i];
				if(sum[i]>9)
				{
					sum[i] -= 10;
					sum[i+1] ++;
				}
			}
		}
		
		int cn = 0;
		for(int i=N-1;i>=0;i--)
		{
			if(sum[i])
			{
				cn = i;
				break;
			}
		}
		for(int j=cn;j>=0;j--)
			printf("%d",sum[j]);
		printf("n");
		if(t!=0)
		{
			printf("n");
		}
	}
}