412 - Pi

[UFP2161]

Random Input:

6
924
11365
5809
10130
14903
28986
0

Should output:

3.000000

But you output:

3.464102

[midra]

thanks a lot UFP2161!!!!!(I don't understand your nick)
I finally got AC!!
I just forgot some stupid part of the problem!!
Thanks again!!!!
wish you good luck!!!
byee

[UFP2161]

If you've ever watched Star Trek, you might get it.. Yes, I know it has absolutely nothing to do w/ the picture.

[jagadish]

tried all those..still WA

can anyone give me critcal IO for this one

[RustB]

Change your gcd function to this

Code: Select all

int gcd(int p,int q)
{
	if (q > p) return(gcd(q,p));
	if (q == 0) return(p);
	return( gcd(q, p % q) );
}

Change the call accordingly.
Then, remove the check for repeated numbers.
Further, get rid of the extra variables x[1] and x[2] and change 6 to 6.0
Then, remove the various increments for total, just change it to
total=n*(n-1)/2;
*whew* That should do it

[Rolando_de_Gilead]

Someone can help me? I can't see where is the problem but the systems says WA.
-------------------------------------------------------------------------------------

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

typedef struct {
int dim;
int v[50];
} set;

int readinput (set f[]) {
int i=0, j=0;

while(f.dim!=0) {

scanf("%d",&f.dim);

if (f.dim!=0) {

for (j=0;j<f.dim;j++) {

scanf("%d",&f.v[j]);}

i++; }}

return i;}



int cmp (int a, int b) {

int i; double r=0;


if (a>b && b!=1) {

for (i=2;i<=b;i++) {

if (a%i==0 && b%i==0 ) {

r= 1; } }
}

else if (b>a && a!=1) {

for (i=2;i<=a;i++) {

if (a%i==0 && b%i==0 ) {

r= 1; } }
}


else if (a==b ) {

r= 1; }


return r;
}


double sum (int n) {

int i, r=0;
for (i=1; i<n;i++) {
r=r+i; }
return r; }


void main() {
int i, max, j, k;
double r2;
set f[50];
double count[50];
double r;
max = readinput (f);

for (i=0; i<max; i++) {
count=0;

for (j=0; j<f.dim; j++) {

for (k=0; k<f.dim; k++) {


if (cmp( f.v[j], f.v[k]) ==0 ){

count [i] ++; } }

}
if (count [i]>0) {
r = sqrt ( 6*(2*sum(f[i].dim)) / (count[i]) );

printf ("%.6lf\n", r); }
else
printf ("No estimate for this data set.\n") ;
}



}

--------------------------------------------------------------------------------------
Thanks

[dovier_antonio]

Tries by this way:

int gcd(int a, int b) {
return (a % b == 0) ? b : gcd(b, a % b);
}

[Raj Ariyan]

Hi,
Please dont post your AC code. You have post lots of your ACC code which is not a good work.

[Sedefcho]

Some sample I/O for anyone who might be interested.

Hope it will be helpful for someone.


INPUT

Code: Select all

5
2
3
4
5
6
2
13
39
3
4
5
6
22
2
4
5
6
90
99
101
102
103
104
105
106
1001
1002
1003
1004
10005
10008
10009
20001
30001
30003
9
32767
32000
16001
15050
1001
1002
100
99
101
11
9999
10001
20000
185
193
202
303
304
305
555
551
5
14
222
770
552
1110
0

OUTPUT

Code: Select all

3.162278
No estimate for this data set.
3.000000
3.113247
2.939388
2.872281
No estimate for this data set.

[Yu Fan]

in the problem we have to estimate the value of pi by evaluating the number of pairs of numbers without common factor.

i computed the pairs and use the fomula problem provides to produce the answer, but i always get WA. my value is like this:

sqrt((3*n*(n+1)/p)),

the n is the size of input and p is the number of pairs.
finally, i eliminate the outer parenthesizes in sqrt() like this

sqrt(3*n*(n+1)/p),

and got AC.
i don't understand why this got AC but the former didn't?
somebody help?

[Hector_Hsu]

Here is my code

Will some function like sqrt() make errors or I miss any part here?

Please tell me , thanks a lot. ^^

Code: Select all

#include<cstdio>
#include<cmath>
using namespace std;
int prime=0;
int judge(int a,int b);
int main(void)
{
    int N,n[50];
    double pi;
    while(scanf("%d",&N)==1)               
    {
        if(N==0)                            
        return 0;
        
        for(int j=0;j<N;j++)                 
        {
            scanf("%d",&n[j]);
        }
        for(int k=0;k<N-1;k++)              
        {
            for(int l=k+1;l<N;l++)
            {
                judge(n[k],n[l]);           
            }
        }
        if(prime==0)                         
        {
            printf("No estimate for this data set.\n");
        }
        else                                 
        { 
            pi = (double)sqrt(3.0*N*(N-1)/(double)(prime+0.0));
            printf("%.6lf\n",pi);
        }
        
        for(int i=0;i<50;i++)                
        {
            n[i] = 0;
        }
        prime = 0;
        pi = 0;
    }
}
int judge(int a,int b)
{
     int c,d;
     if(a==b)
     {
         return 0;
     }
     else if(a>=b)
     {
         c = a;
         d = b;
     }
     else if(b>a)
     {
         c = b;
         d = a;
     }
     
     int r = c%d;
     if(r==1)
     {
         prime++;
         return 0;
     }
     else if(r==0)
     {
         return 0;
     }
     else
     {
         c = d;
         d = r;
         judge(c,d);
     }
}

[Hector_Hsu]

Sorry,

the title is find the "bug" , not the "big" |||

/________________________\

[jaracz]

I did it like this

Code: Select all

double pairs = n*(n-1)/2;
double pi = sqrt(6*pairs/count);

where pairs are the maximum quantity of pairs and count is amount;) of pairs in which greatest common divisor is 1

and your function may be done much easier(actually I dont understand what you've tried to do there, it could be as well as void, couse it returns value to nothin', but your prog succeded in/out, so nevermind)

Code: Select all

int judge(int a,int b) // gcd algo...
{ 
     while(a != b) if(a > b) a -= b; else b -= a;
     return(a); 
}

(your "prime" is my "count")

Regards

[jaracz]

and sqrt() in <cmath> sometimes calculate incorrect answers(dont know why),so you'd better not to declare "namespace" standard, just change it to <math.h> instead of <cmath> and remove "using namespace std;"
(stdio also)

[=viki=]

why TLE plz help,,,

time limit exceeded

Code: Select all

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

int _fact(long int a,long int b)
{
	long int i,j;
	for(i=2;i<=a || i<=b;i++)
	{
		if(a%i==0 && b%i==0)
			return 1;
	}
	return 0;
}

int main(void)
{

	long int N,count=0,total=0,i,j;
	long int arr[50];
	while(1)
	{
		scanf("%ld",&N);
		if(N==0)
			break;
		for(i=0;i<N;i++)
			scanf("%ld",&arr[i]);
		for(i=0;i<N-1;i++)
		{
			for(j=i+1;j<N;j++)
			{
				total++;
				if(!_fact(arr[i],arr[j]))
					count++;
			}
		}
/*	       	printf("%ld %ld\n",total,count);    */
		if(count==0)
			printf("No estimate for this data set.\n");
		else
			printf("%.6f\n",sqrt((total*6)/count));
		count=0;
		total=0;
	}
	return 0;
}

i itried all the inputs pz help....