10656 - Maximum Sum (II)

[sohel]

Definition of subsequence : subsequence is a term used to mean a sequence derived from a sequence S by discarding some of its terms.
-- from mathworld

So, in the above example 2 5 is a subsequence of the given sequence.

[CodeMaker]

hmm, thanks... looks like I have problems in basic

[Stummer]

Thanks to Adrian Kuegel !!! At last I got Accepted!

[bishop]

why this is wa
is there any critical way to solve
this

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[darkos32]

i got PE..can someone help me give me the sample input and output...

this is my code :

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got accepted..

thanks..

[alamgir kabir]

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//remove

Thanks Jan for helping me.

[Jan]

Check the set very carefully.

Input:

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6
1
0
3
4
0
1
0

Output:

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1 3 4 1

Hope you get accepted.

[alamgir kabir]

I do not consider this case.
Now I have acc it.
Thanks a lot Jan. I am greatful to u.

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//code removed

Thanks for helping me.
I hope u wil help me in future.

[Jan]

Try the case.

Input:

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3
0
0
0
0

Output:

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0

Hope it helps.

[kbr_iut]

now AC, code removed

[asif.mist]

is there any trailing space??
i got 5 PE....

[aaa111]

I am getting PE,please help:

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#include<stdio.h>

#define SIZE 1010

int main(void)
{
	int i,j,n,flag=0;
	unsigned int num[SIZE];

	while(1)
			{
			flag=0;
			scanf("%d",&n);
			if(n==0)
				break;
			for(i=0;i<n;i++)
				scanf("%u",&num[i]);
			
			if(num[0]==0)
				flag=1;

			i=0;
			if(flag==1)
				while((num[i]==0) && (i<n))
					i++;
					
			if(i==n)
				 printf("0");
			else
				{
					for(j=i;j<n;j++)
						{
							if(num[j]!=0)
								{
								printf("%u",num[j]);
								if(j!=(n-1))
									printf(" ");
								}
						}
				}
	
			printf("\n");
			}
	
	return 0;
}

[plamplam]

Very easy problem
As the numbers are all non-negative, when you are taking input just store the number in an array if the number is not equal to 0. And later just print the array(with no trailing spaces).

[blckhrt]

pls help..um continuosly getting wa in this code..wat's wrong???


#include<stdio.h>
int main()
{
long long int n,i,j,k,flag,num[10000];
while(scanf("%lld",&n)==1)
{
if(n==0)
break;
flag=0;
for(i=0;i<n;i++)
{
scanf("%lld",&num);
if(num==0)
{
flag++;
}
}

if(flag==n)
printf("0\n");
else
{
for(j=0;j<n;j++)
{
if(num[j]==0)
continue;
printf("%lld ",num[j]);
}

printf("\n");
}

}
return 0;
}

[kat]

Please Check this input/output:

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6
3 4 0 3 4 5
5
1 0 0 0 2
3
0 0 0
4
0 0 1 2

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3 4 3 4 5
1 2
0
1 2