382 - Perfection

[jagadish]

read the problem:
Format counts: the echoed integers should be right justified within the first 5 spaces of the output line, followed by two blank spaces, followed by the description of the integer. The final line of output should read END OF OUTPUT.

[efr_shovo]

It is My Resolve Code.But Still P.E

#include<stdio.h>

long count,sum,n[100],d,mod,m,z;

int input()
{
int i=0,f;
do
{
scanf("%d",&n);
f=n;
i++;
}while(f!=0);
return i;
}

void space(int j)
{
if(n[j]<10)
printf(" ");
if(n[j]>=10&&n[j]<100)
printf(" ");
if(n[j]>=100&&n[j]<1000)
printf(" ");
if(n[j]>=1000&&n[j]<10000)
printf(" ");
if(n[j]>10000)
printf(" ");
}

void main()
{
int i=0;
z=input();
if(z>1&&z<100)
{
printf("PERFECTION OUTPUT\n");
while(1)
{
if(n>60000)
break;
if(n==0)
break;
d=n/2;
for(count=1;count<=d;count++)
{
mod=n%count;
if(mod==0)
sum=sum+count;
}
space(i);
if(sum>n)
printf("%ld ABUNDANT\n",n);
else
if(sum==n)
printf("%ld PERFECT\n",n);
else
printf("%ld DEFICIENT\n",n[i]);
sum=0;
i++;
}
printf("END OF OUTPUT");
}
}

[jagadish]

there are easier ways to format..
[c]
#include<stdio.h>

long count,sum,n[100],d,mod,m,z;

int input()
{
int i=0,f;
do
{
scanf("%d",&n);
f=n;
i++;
}while(f!=0);
return i;
}

void main()
{
int i=0;
z=input();
if(z>1&&z<100)
{
printf("PERFECTION OUTPUT\n");
while(1)
{
if(n>60000)
break;
if(n==0)
break;
d=n/2;
for(count=1;count<=d;count++)
{
mod=n%count;
if(mod==0)
sum=sum+count;
}

if(sum>n)
printf("%5ld ABUNDANT\n",n);
else
if(sum==n)
printf("%5ld PERFECT\n",n);
else
printf("%5ld DEFICIENT\n",n[i]);
sum=0;
i++;
}
printf("END OF OUTPUT\n");
}
}[/c]

[jaracz]

hi guys!

It is necessity to read in input to array?
Can I read one integer(long) and write answer, then read next and so on and on??

It is possible that input may exceed 100 and then program shouldn't print anything??

w8ing 4reply..thx in advance

[Jemerson]

I'm not so sure about this, but i guess your read buffer wont hold about 100 integer like 59999, will it? U'd better increase it and try again.

[Jemerson]

I found the mistake, the input is in multiple lines, read all lines and process until find the 0, I don't know if there are empty lines, so.. you'd better take care of this too.

Good Luck

[thinker_bd]

PLEASE SAY ME WHAT IS MY WRONG
WHY I GETTING WA. MY CODE IS GIVEN BELOW

Code: Select all

#include<stdio.h>
int main()
{
	long n,i;
	printf("PERFECTION OUTPUT\n");
	freopen("328.txt","r",stdin);
	while(1)
	{
		scanf("%d",&n);
		if(n==0) 
		{
			printf("END OF OUTPUT\n");
			break;
		}	
		long sum;
		for(i=1;i<=n/2;i++)
		{
			if(i==1)sum=1;
			else
			{
				if((n%i)==0)
					sum+=i;
			}
		}
		
		long count =0,p=n;	
			while(1)
			{
			if(p==0) break;
			p=p/10;
			count++;
			}
	

		printf("%5d",n);
		printf(" "); //2space
		
		if(sum>n)
			printf("ABUNDANT\n");
		
		else if(sum==n)
			printf("PERFECT\n");
		
		else if(sum<n)
			printf("DEFICIENT\n");
	  }
	return 0;
}

[neno_uci]

1 is DEFICIENT

[mars kaseijin]

Greetings thinker_bd,
here are some recommendations:
* avoid file i/o as Judge supplies its own set of inputs.
* try to de-couple the algorithm from point of application. It is cleaner design and more functional.

Thanks for reading

[abhi]

i get WA for my code but it passes all the sample test cases

here it is.........

Code: Select all

DELETED after AC

[helloneo]

I don't think your program print "END OF OUTPUT" at the end of output..

[abhi]

oops!!! silly me ......

[athlon19831]

i don't know why i got WA instead of AC.
my code following:
#include <iostream>
using namespace std;
#include <cstring>
#include <cstdio>
#include <cmath>
int main(int argc, char* argv[])
{
char str[1000];
long copy[110];
long s[110];
long i,j,k,l,sum,val,i_save,j_copy;
//bool flag;
for(i=0;i<100;i++)
{
s=0;
}
gets(str);
l=strlen(str);
sum=0;val=10;j=0;
for(i=0;i<l;i++)
{
if(str>='0'&&str<='9')
{
sum=sum*val+int(str)-48;
}
else
{
copy[j]=sum;
j++;
sum=0;
}
if(copy[j-1]=='0')
break;
}
j_copy=j;
for(i=0;i<j_copy;i++)
{
i_save=i;

for(j=2;j<=int(sqrt(copy));j++)
{
if(copy%j==0)
{
i_save=copy/j;
s=s+i_save+j;
}
}
s=s[i]+1;

}

cout<<"PERFECTION OUTPUT"<<endl;
for(i=0;i<j_copy;i++)
{
if(s[i]==copy[i])
{
printf("%5d",copy[i]);
cout<<" PERFECT"<<endl;
}
else if(s[i]<copy[i])
{
printf("%5d",copy[i]);
cout<<" DEFICIENT"<<endl;
}
else
{
printf("%5d",copy[i]);
cout<<" ABUNDANT"<<endl;
}
}
cout<<"END OF OUTPUT"<<endl;

return 0;
}

[athlon19831]

382 WA pls give me some test cases

[raygun]

I noticed that you have a presentation error - there are supposed to be two spaces. Even if you fix the program bug, you'll still miss out.

Code: Select all

	gets(str);
	l=strlen(str);
	sum=0;val=10;j=0;
	for(i=0;i<l;i++)
	{
		if(str[i]>='0'&&str[i]<='9')
		{
		   sum=sum*val+int(str[i])-48;
		}
		else
		{
			copy[j]=sum;
			j++;
			sum=0;
		}
		if(copy[j-1]=='0')
			break;	
	}

This is the first problem right here - you assume the input is on one line. Also, the implementation in question chops off the last integer on that line as well since it doesn't even check the null character at the end to see if the number needs to be added.

Code: Select all

s[i]=s[i]+1; 

This is the second problem - it incorrectly makes 1 a perfect number.