Page 9 of 14
Posted: Sun Jul 22, 2007 11:32 am
by helloneo
Read the previous posts.. (about EPS)
Posted: Tue Jul 31, 2007 5:21 pm
by ankit.arora
I m gettin WA but i have checked my output for all inputs provided in this thread!..... Please help me!!!
Code: Select all
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
unsigned long long i,j,count,t;
int set;
int arr[100001]={0};
arr[0]=1;
arr[1]=1;
for(i=2;i<100001;i++)
{
if(arr[i]==0)
{
j=2;
while(i*j<100001)
{
arr[i*j]=1;
j++;
}
}
}
unsigned long long num[10001];
count=0;
for(i=0;i<=10000;i++)
{
t=(i*i)+i+41;
if(t<=100000)
{
if(arr[t]==0)
count++;
num[i]=count;
}
else
{
set=0;
for(j=3;j<=sqrt((double)t+0.5);j+=2)
{
if(t%j==0)
{
set=1;
break;
}
}
if(set==0)
count++;
num[i]=count;
}
}
int a,b;
double per;
while(cin>>a>>b)
{
if(a==0)
per=100*((double)(num[b])/(double)(b-a+1));
else
per=100*((double)(num[b]-num[a-1])/(double)(b-a+1));
t=(long long)(per*100);
if((per*100)-t>0.5 ||((per*100)-t==0.5 && t%2!=0))
t++;
if(t%100==0)
cout<<(double)t/(double)100<<".00\n";
else if(t%10==0)
cout<<(double)t/(double)100<<"0\n";
else
cout<<(double)t/(double)100<<"\n";
}
}
Posted: Wed Aug 01, 2007 4:11 am
by mmonish
>>ankit.arora
did u check the sample inputs. for the following input 1423 2222 ur output is 44.12 but the correct output is 44.13.
use double calculation and output like this
printf(".2lf\n" , res);
and don't forget about EPS.
Hope this helps.
Posted: Wed Aug 01, 2007 7:41 pm
by ankit.arora
mmonish wrote:>>ankit.arora
did u check the sample inputs. for the following input 1423 2222 ur output is 44.12 but the correct output is 44.13.
use double calculation and output like this
printf(".2lf\n" , res);
and don't forget about EPS.
Hope this helps.
Actually for this input the actual floating answer is 44.125(as per my program) which when rounded to two decimal places produces 44.12, this is the rule which i use in all my rounding calculations ..... can you please explain me that why it should be 44.13 instead.
Thanks a lot for helping!!!!
Posted: Tue Aug 07, 2007 5:59 am
by shaiful
I got 20 WA in this problem
Can any one help me on my code?
Code: Select all
#include<stdio.h>
#include<math.h>
bool prime(long long n)
{
long long i,sq;
sq=(long long)sqrt((double)n);
for(i=2;i<=sq;i++)
{
if(n%i==0) return false;
}
return true;
}
int main()
{
long long dif,count,temp,t,i,j,n1,n2,k;
bool p[10005];
for(j=0;j<10005;j++) p[j]=false;
for(j=0;j<10005;j++)
{
temp=j;
k=temp*temp+temp+41;
if(prime(k)==true)
p[j]=true;
}
double per;
while(scanf("%lld %lld",&n1,&n2)==2)
{
count=0;
if(n1>n2)
{
t=n1;
n1=n2;
n2=t;
}
for(i=n1;i<=n2;i++)
{
if(p[i]==true) count++;
}
dif=n2-n1+1;
per=(double)((double)count/(double)dif)*100.0;
printf("%.2lf\n",per);
}
return 0;
}
Posted: Tue Aug 07, 2007 6:08 am
by emotional blind
This problem can be solved without using float or double..
That will be more safe.
Posted: Thu Aug 09, 2007 4:21 pm
by WingletE
I've passed all the test case above but still got WA, I use long double.
What could the problem be?
This problem can be solved without using float or double..
then how?
Posted: Thu Aug 09, 2007 4:39 pm
by mf
You could round and output a fraction p/q, for example with this code:
Code: Select all
int t = (p * 200 + q) / (2 * q);
printf("%d.%.2d\n", t / 100, t % 100);
WA(10200)[precision error]
Posted: Sun Aug 19, 2007 3:42 pm
by ishtiaq ahmed
Posted: Sun Aug 19, 2007 4:12 pm
by WingletE
I've run your program. Your answer is not right.
Code: Select all
9999 10000
9998 10000
9998 9999
9996 10000
500 600
post reply of 10220 [prime time]
Posted: Sun Aug 19, 2007 7:16 pm
by ishtiaq ahmed
i rewrite my code as follows but stll facing WA.
waiting for your reply
....
Posted: Sun Sep 23, 2007 7:33 pm
by Fuad Hassan EWU
WA!!! i can't find the bug. i used very simple and i think simple logic coz this seems to be eay problem. but where is the bug? I matched the provided critical input output. plz help...
Code: Select all
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<ctype.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#define N 120000000
using namespace std;
char a[N/2+10];
void genprime()
{
long i,j;
for(i=3;i*i<=N;i+=2)
{
if(a[i/2]==1)
continue;
for(j=(i*i)/2;j<=N/2;j+=i)
{
a[j]=1;
}
}
}
int main()
{
//freopen("input.txt", "r", stdin);
//freopen("output.txt", "w", stdout);
genprime();
double prmcnt;
double result;
long i,first,b, f;
while(cin>>first>>b)
{
prmcnt=0;
for(i=first;i<=b;i++)
{
f=(i*i)+i+41;
if(f==2)
prmcnt++;
else if(f%2==0&&f!=2)
continue;
else
{
if(a[f/2]==0)
prmcnt++;
else
continue;
}
}
result=(100*(double)prmcnt)/(double)((b-first)+1);
printf("%.2lf\n",result);
}
return 0;
}
Posted: Sun Sep 23, 2007 8:11 pm
by Jan
Your code is taking too much memory.
...
Posted: Tue Sep 25, 2007 2:20 pm
by Fuad Hassan EWU
jan vai i used the size 100010041, because 10000^10000+10000+41=100010041. and to check whether it is prime or not i need to run a the outer of my seive method up to sqrt(100010041). am i right? plz make me clear about this issue. thanks.
Posted: Tue Sep 25, 2007 5:30 pm
by Jan
int (sqrt(100010041)) = 10000. There are less than 2460 primes in [2,10000]. So, in worst case you have to divide 2460 times to check whether a number is prime or not. And this isn't too much. It uses less memory, but takes a little more time. So, you can generate primes upto 10000, and make your primarity testing. Hope it helps.