412 - Pi

[ram]

My code works Ok with all the test cases. But I am getting "WA".Are there any special cases??

Code: Select all

#include <iostream>
#include <math.h>

using namespace std;

int isCom(int x,int y)
{
if(y==0) return x;
return isCom(y,x%y);
}

int main(){
int i,j,num;
int cNums;
int nums[51];

while(cin>>num)
{
if(num==0) break;
for(i=0;i<num;i++)
cin>>nums[i];

cNums=0;

for(i=num-1;i>=0;i--)
for(j=i-1;j>=0;j--)
{
	//cout<<nums[i]<<" "<<nums[j]<<" "<<isCom(nums[i],nums[j])<<endl;
	if(isCom(nums[i],nums[j])==1) cNums++;
}


num=num*(num-1)/2;
//cout<<num<<"*6/"<<cNums<<endl;
if(cNums==0)
cout<<"No estimate for this data set."<<endl;
else
cout<<(sqrt((double)(num*6)/cNums))<<endl;
}
return 0;
}

[C8H10N4O2]

Read the question. Round to 6 places:)

[ram]

I fixed it. But still getting WA

[htl]

Why does this code get WA?
[c]
#include<stdio.h>
#include<math.h>
int gcd(int,int);
void main(void)
{
int num[50],n,x,y,total,check;
while(1)
{
scanf("%d",&n);
if(n==0)
break;
for(x=0;x<n;x++)
scanf("%d",&num[x]);
total=n*(n-1)/2;
check=0;
for(x=0;x<n;x++)
for(y=x+1;y<n;y++)
if(gcd(num[x],num[y])==1)
check++;
if(check==0)
printf("No estimate for this data set.\n");
else
printf("%.6lf\n",sqrt(6*(double)total/(double)check));
}
}
int gcd(int x,int y)
{
int a,b,r;
a=x,b=y;
while(1)
{
r=a%b;
if(r==0)
return a;
if(r==1)
return 1;
a=b;
b=r;
}
}
[/c]

[ithamar]

I think that the cause of the WA is that in this part



[c]
for(x=0;x<n;x++)
for(y=x+1;y<n;y++)
if(gcd(num[x],num[y])==1)
check++;
[/c]

You are calculating the MCD of the last number with some UNKNOWN number. thats because the 1st for must end when x < n - 1.

Hope this can help u.

[htl]

But when x=n-1, then y=n and it won't pass the 'y<n' expression and terminate the loop, doesn't it?

[ithamar]

But when x=n-1, then y=n and it won't pass the 'y<n' expression and terminate the loop, doesn't it?

Yep. You are right. My mistake

Another thing that could be is that the numbers are not sorted and in your mcd function you suposse that x >= y. Maybe you are trying to calculate MCD(3, 6) and in your mcd this is not the correct answer but MCD(6, 3) it is


Hope this can help u

[htl]

Is it possible that the same number appears more than once? And I think the gcd() will execute extra one time if x<y.

[imranul]

Dear htl....

I think the problem is in the gcd function....

suppose you are going to calculate gcd(3,1)
then x=3,y=1 and as a=x and b=y a=3 and b=1

So, in the loop r=a%b=3%1 (!) and gcd returns a=3
But gcd(3,1) should be 1, isn't it?

[Shahid]

hi,

isn't it possible to do without using a GCD funtion??? i tries in the following way and the result is WA. here is my code :


[cpp]
/*@BEGIN_OF_SOURCE_CODE*/


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

void main()
{
int N, in[55], k, flag = 0, n, tmp, res;
register int i, j, count = 0;
long double pi;

for(;;)
{
scanf("%d", &N);

if(!N)
break;

for(i = 0; i < N; i++)
scanf("%d", &in);

count = 0;
n = (N * (N-1))/2;
for(i = 0; i < N; i++)
for(j = i+1; j < N; j++)
{
flag = 0;
if(!(in%in[j]) || !(in[j]%in) || (!(in%2) && !(in[j]%2)))
flag = 1;

if(!flag)
{
for(k = 2; k <= sqrt(in); k++)
{
tmp = in%k;

if(!tmp)
{
res = in/k;

if(!(in[j]%k) || !(in[j]%res))
{
flag = 1;
break;
}
}
}
}
if(!flag)
count++;
}

if(count)
{
pi = sqrt((6 * n)/count);
printf("%.6Lf\n", pi);
}
else
printf("No estimate for this data set.\n");
}
}
/*@END_OF_SOURCE_CODE*/
[/cpp]

[razibcse]

Please someone inform me abot my coding of this problem...

Code: Select all

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

void main()
{
int t,p,n,a,b,c,nm[50],i,j,u,l;
float pis,d;
for(;;)
{
scanf("%d",&n);
if(n==0) break;
for(a=0;a<n;a++)
 scanf("%d",&nm[a]);
u=0;
l=0;
for(a=0;a<(n-1);a++)
 for(b=a+1;b<n;b++)
  {
  p=1;
  l++;
  if(nm[a]>nm[b])
   {
   t=nm[a];
   nm[a]=nm[b];
   nm[b]=t;
   }
  i=nm[a];
  j=nm[b];
  for(c=2;c<=i;c++)
   if((i%c==0) && (j%c==0))
    {
    p=0;
    break;
    }

   if(p)u++;
  }
 if(!u)
  {
  printf("No estimate for this data set.");
  printf("\n");
  continue;
  }
d=(float)l/u;
pis=6*d;
printf("%.6f\n",sqrt(pis));
}

}

[MAXX^FACTOR]

the problem insist that
answers must be rounded to six digits after the decimal point.

If the answer we have is 2.44949
what value we have to output?
2.44949 or 2.449490?

And I don't know why W.A occurs at all...

Code: Select all

#include <iostream.h>
#include <math.h>
#include <iomanip.h>

int GCD(int a,int b){
	while(1){
		if(a<b)
			a^=b^=a^=b;
		if(a%b==0)
			return b;
		a%=b;
	}
}

int C(int x,int y){
	int i,j,mo[100],so[100],t,t2;
	if(x-y>=y)
		t=x-y;
	else 
		t=y;
	for(i=0,j=2;i<x-t-1;i++,j++)
		so[i]=j;
	for(i=0,j=t+1;i<x-t;i++,j++)
		mo[i]=j;
	for(i=0;i<x-t-1;i++)
		for(j=0;j<x-t;j++){
			if(so[i]==1)
				break;
			t2=GCD(so[i],mo[j]);
			so[i]/=t2;
			mo[j]/=t2;
		}
	t2=1;
	for(i=0;i<x-t;i++)
		t2*=mo[i];
	return t2;
}

int main()
{
	int i,j,num,seq[10000],S,M,tmp;
	double PI;
	while(1){
		cin>>num;
		if(num==0)
			break;
		for(i=0;i<num;i++)
			cin>>seq[i];
		M=C(num,2);
		for(i=0,S=0;i<num-1;i++)
			for(j=i+1;j<num;j++)
				if(GCD(seq[i],seq[j])==1)
					S++;
		if(S==0){
			cout<<"No estimate for this data set.\n";
			continue;
		}
		PI=(double)sqrt(6*(double)M/(double)S);
	}
	return 0;
}

[Dominik Michniewski]

In your code I couldn't find a final print od PI ) It's not good ....
Answer to your question:

If you have to print output value with 6 digits after decimal point, you must print 6 DIGITS (even if it's six zeroes ) .....

Dominik

[Skywalker1982]

i had the same mistake with float numbers.
when i decided to use double instead of float, it suddenly worked.

[takumi]

i don't know why my code get WA
can anybody tell me why?
[c]
#include <stdio.h>
#include <math.h>
int test ( int a, int b ) {
for( ; ; ) {
if ( a < b )
a^=b^=a^=b;
if ( a % b == 0 )
return b;
a %= b;
}
}
int main()
{
int input, a[100], i, j, counter = 0, total;
double temp, temp2;

for ( ; ; ) {
scanf("%d", &input);
counter = 0;
if ( input == 0 )
break;

for ( i = 0; i < input; i++ ) {
scanf("%d", &a);
}
total = input * ( input - 1 ) / 2;

for ( i = 0; i < input - 1 ; i++ ) {
for ( j = i+1; j < input; j++ ) {
if ( a == a[j] )
break;
if ( test( a,a[j] ) == 1 ){
counter++;
}
}
}
if ( counter == 0 ){
printf("No estimate for this data set.\n");
continue;
}

temp = 6 /( double)counter;
temp2 = sqrt(total * temp);
printf("%.6f\n", temp2);
}
return 0;
}[/c]
thanks..