1180 - Perfect Numbers
Posted: Thu Apr 03, 2014 3:00 pm
On this problem, note that the integers on the second line are separated by commas - and not spaces.
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Re: 1180 - Perfect Numbers
Posted: Sat Aug 02, 2014 12:24 pm
Re: 1180 - Perfect Numbers
Posted: Sat Aug 02, 2014 3:24 pm
Why do you think that p is one digit number? It can be greater than 9.
This problem is tricky. I checked that p can be greater than 1000.
So you can forget about checking if sum of proper divisors are equal to a number.
And you don't need to check it.
Think about this condition
Code: Select all
p=num[i]-'0';
n=pow(2,p-1)*(pow(2,p)-1);So you can forget about checking if sum of proper divisors are equal to a number.
And you don't need to check it.
Think about this condition
And this conditionEuclid proved that an even number is perfect if it has the form
2^(p-1) * (2^p - 1)
where both p and 2^p - 1 are prime numbers.
By the way i read p this wayThe largest perfect number in this problem will not exceed 2^33
Code: Select all
scanf("%d", &t);
while (t--) {
scanf("%d", &p);
if (t) scanf(" %c", &c);
..
}
Re: 1180 - Perfect Numbers
Posted: Sat Aug 02, 2014 10:11 pm
If max value of perfect number is 2^33.
Then 2^(p-1) * (2^p - 1) <= 2^33. Can you get max value for p?
If given p is greater than max value i print "No" because they will be not perfect numbers according to problem description.
Then 2^(p-1) * (2^p - 1) <= 2^33. Can you get max value for p?
If given p is greater than max value i print "No" because they will be not perfect numbers according to problem description.
Re: 1180 - Perfect Numbers
Posted: Sun Aug 03, 2014 4:21 pm
Your prime checking is wrong.
When you solve problems relating to prime numbers -> Add a condition that 1 is not a prime.
(if problem description didn't said anything about it)
It would be better if you make function isPrime(int n) that checks if number is prime.
When you solve problems relating to prime numbers -> Add a condition that 1 is not a prime.
(if problem description didn't said anything about it)
It would be better if you make function isPrime(int n) that checks if number is prime.
Re: 1180 - Perfect Numbers
Posted: Fri Nov 28, 2014 9:03 pm
getting TL
any suggestion ?
here is my code
any suggestion ?
here is my code
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#include<stdio.h>
#include<math.h>
int main()
{
unsigned long long int n,i,j,p,sum,t;
//freopen("1180.txt","r",stdin);
while(scanf("%llu",&t)==1)
{
for(i=0;i<t;i++)
{
scanf("%llu",&p);
if(i<t-1)
scanf(",");
sum=0,n=0;
n=pow((double)2,p-1)*(pow((double)2,p)-1);
for(j=1;j<n;j++)
{
if(n%j==0)
sum+=j;
}
if(n==sum)
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}Re: 1180 - Perfect Numbers
Posted: Fri Nov 28, 2014 9:20 pm
Helaluddin_brur,
According to your code, you should print Yes, when both f and f1 is 0.
According to your code, you should print Yes, when both f and f1 is 0.
Re: 1180 - Perfect Numbers
Posted: Sat Nov 29, 2014 1:04 pm
Doesn't match sample I/O. Input numbers are separated by commas(','). You should read this commas from standart input.
Re: 1180 - Perfect Numbers
Posted: Thu Dec 04, 2014 10:04 pm
Why I am getting TLE ?? Please Help ....
Code: Select all
#include<stdio.h>
#include<math.h>
int main()
{
long long int n, i, p, a[100000], j, d, sum;
char ch;
while(scanf("%lld",&n)==1)
{
for(i=1;i<=n;i++)
{
scanf("%lld,",&a[i]);
scanf("%c",&ch);
}
for(i=1;i<=n;i++)
{
d = (pow(2,a[i]-1))*(pow(2,a[i])-1);
sum=0;
for(j=1;j<d;j++)
{
if((d%j)==0)
sum = sum+j;
}
if(sum==d)
printf("Yes\n");
else
printf("No\n");
}
}
return 0;
}
Re: 1180 - Perfect Numbers
Posted: Fri Dec 05, 2014 12:37 am
Re: 1180 - Perfect Numbers
Posted: Fri Dec 05, 2014 11:44 am
ssavi, if you want to optimize your code read posts in this thread.