412 - Pi

[Jan]

Try the set.

Input:

Code: Select all

6
1
31727
490
407
677
29070
0

Output:

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2.535463

Hope it helps.

[rhsumon]

Where i the problem here
I can't fix
Plz take a look

Code: Select all

Silly Mitake 
Anyway AC now

Plz help anyone

[pok]

Why WA ?
cant find any mistake..
pls someone help me..

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removed after AC..

need ur help..

[Articuno]

Check this line:

Answers must be rounded to six digits after the decimal point.

Good luck

[pok]

Thanks a lot..
Take care.. & God bless you..

[Jordi Aranda]

I don't pass the proposed inputs in this thread but I don't know what I'm doing wrong. Here's my code:

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Removed after accepted

It seems that for large inputs, something is wrong with my code and gives a wrong expected pi.

[mf]

In this line:

double expected_pi = sqrt(6*npairs/compta);

you have an integer division, which rounds the result of division down to the nearest integer.

Use something like:

Code: Select all

double expected_pi = sqrt(6*npairs/(double)compta);
  or:
double expected_pi = sqrt(6.0*npairs/compta);

[Jordi Aranda]

Thanks again mf

[kawsar]

#include<stdio.h>
#include<math.h>

long a[1000000];

int main(){
long n, i, j, x, p, q ;

while(scanf("%ld",&n)==1){
long num =0, sum=0;
if(n == 0)
return 0;

for(i=0; i<n; i++)
scanf("%ld",&a);


for(i=0; i<n; i++){
for(j=i+1; j<n; j++){
num++;

if(a == 1 && a[j]==1){
sum++;
continue;
}


x = a % a[j];
p = a[j];

while(x != 0){
if(x == 1){
sum++ ;
break ;
}
q = x ;
x = p % x ;
p = q ;
}

}
}

if(sum == 0)
printf("No estimate for this data set.\n");
else
printf("%.6f\n",sqrt((6/(float)sum)*num));
}
return 0;
}


//In every time i got wrong answer. why ??

[kawsar]

I found every time Wrong Answer for this problem...
Is there any spacial input output for this problem ???

If there , then any one please give me the input output.... I am so fucked up for this problem...
Help me...........!!!

[shatil_cse]

please help me ....
don't know why WA!!!!!!!!!!

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//code removed after AC

[brianfry713]

On my machine your code prints 0.000000 for the first sample input. Your printf is using %Lf or long double, but sqrt returns a %lf or double.

[shatil_cse]

On my machine your code prints 0.000000 for the first sample input. Your printf is using %Lf or long double, but sqrt returns a %lf or double.

Oh got AC ........
So thanks bosssssss

[tansif.ruetcse13]

Code: Select all

#include<stdio.h>
#include<math.h>
int main()
{
    int i,j,a[50],d,temp,N,t;
    while(1)
    {
    d=0,t=0;
    scanf("%d",&N);
    if(N==0)
    break;
    for(i=0;i<N;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<N-1;i++)
    {
        for(j=i+1;j<N;j++)
        {
            if(a[i]<a[j])
            {
                temp=a[i];
                a[i]=a[j];
                a[j]=temp;
            }
        }
    }
    for(i=0;i<N-1;i++)
    {
        for(j=i+1;j<N;j++)
        {
            if((a[i]%a[j]!=0))
            {
                if((a[i]%2!=0)||(a[j]%2!=0))
                d++;
            }
            t++;
        }
    }
    if(d==0)
    printf("No estimate for this data set.\n");
    else
    printf("%.6lf\n",sqrt((6*t)/d));
}
return 0;
}

why wa??

[tansif.ruetcse13]

Code: Select all

#include<stdio.h>
#include<math.h>
int main()
{
    int i,j,a[50],d,temp,N,t;
    while(1)
    {
    d=0,t=0;
    scanf("%d",&N);
    if(N==0)
    break;
    for(i=0;i<N;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<N-1;i++)
    {
        for(j=i+1;j<N;j++)
        {
            if(a[i]<a[j])
            {
                temp=a[i];
                a[i]=a[j];
                a[j]=temp;
            }
        }
    }
    for(i=0;i<N-1;i++)
    {
        for(j=i+1;j<N;j++)
        {
            if((a[i]%a[j]!=0))
            {
                if((a[i]%2!=0)||(a[j]%2!=0))
                d++;
            }
            t++;
        }
    }
    if(d==0)
    printf("No estimate for this data set.\n");
    else
    printf("%.6lf\n",sqrt((6*t)/d));
}
return 0;
}