OJ Board

Try this input:
1 0.9

The output will be: 1 / 1 ----------> WRONG ANSWER!!!

No, that is correct. We always assume 0.9999999... to be equal to 1.

Bye.
Andrey.

could somebody tells my how use better precison in this problese. i got TLE but my prog and algorithm work fine, but somthing w precision. Please help.
Here is my source code C++
[cpp]

#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <fcntl.h>
#include <iostream>
using namespace std;

long int euklid (long int x,long int y)
{
long int g=1,t;
while ((x%2==0) && (y%2==0))
{
x/=2;
y/=2;
g*=2;
}
while (x>0)
{
if (x%2==0)
x/=2;
else if (y%2==0)
y/=2;
else
{
t=abs(x-y)/2;
if (x>=y)
x=t;
else
y=t;
}
}
return g*y;
}

long double pot(int pod,int wyk)
{
int wsk,tmp=pod;
if (wyk==0)
return 1;
for (wsk=2;wsk<=wyk;++wsk)
pod*=tmp;
return (long double)(pod);
}

int main()
{
#ifndef ONLINE_JUDGE
close (0); open ("myprog.in", O_RDONLY);
close (1); open ("myprog.out", O_WRONLY | O_CREAT, 0600);
#endif

long double ulamek,tmp;
long int p,q,podz;
int i,j,wsk,liczba,cas=1;

while (cin >> j >> ulamek)
{
cout << "Case " << cas << ": ";
++cas;
i=0;
tmp=ulamek;
while (tmp-(long double)((int)(tmp+0.0000000001))>=0.0000000001) // problem is there beacuse i get (int)((long double)(30))=29
{
++i;
tmp*=(long double)(10);
}
i-=j;
p=(long int)(pot(10,i+j)*(ulamek+0.00000000001))-(long int)(pot(10,i)*(ulamek+0.00000000001)); //or problem is there;
q=(long int)(pot(10,i+j)-pot(10,i));
podz=euklid(p,q);
p/=podz;
q/=podz;
cout << p << '/' << q << '\n';
}
}
[/cpp]
best regards

You don't need to use floating point arithmetic for this. Integers work and are not subject to such errors (unless you overflow of course) =)

I got RTE for the following code: What is the wrong with this code?

why don't you post it in volume 3 ?

The above, and be careful of input like:

1 0.0

i first submitted my code and got RTE (floating point error) as i did not handled the case j=0 . now that i have handled it , i am getting WA . can some one get me some critical IO for the prob .

i tried to solve the prob as per as the statement of the prob .

Bye
Riyad

There's a post from May 14, 2002 with test data by Ivan Golubev.

But they're probably insufficient as my program has not been accepted.

My code got WA. Please help me to find the bug in my code:-
[c]#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>

long gcd(long a, long b)
{
if(!b) return a;
else return gcd(b, a%b);
}

void main()
{
int j, k, len, kase=1;
long num, denom, g;
double x, a, b;
char ara[20];

while(1==scanf("%d", &j) && j!=-1)
{
scanf("%s", ara);
len = strlen(ara);
k = len - j - 2;
x = atof(ara);

a = pow(10, k+j) * x;
a = floor(a + .005);
b = pow(10, k) * x;
b = floor(b + 0.0000000005);

num = a - b;
denom = pow(10, k+j) - pow(10, k);

g = gcd(num, denom);

if(g)
{
num /= g;
denom /= g;
}

printf("Case %d: %ld/%ld\n", kase++, num, denom);
}
}[/c]

here's my code:
[cpp]
#include <iostream.h>
#include <math.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
int main()
{
int a,b,i,d,j,k,m,cases=0,y;
char c,p[100];
while(cin>>j,j!=-1)
{cases++;
cin>>c>>c;
cin>>p;
k=strlen(p);
k-=j;
b=atoi(p);

i=1;
for(m=0;m<j;m++)i*=10;
d=b/i;

i=1;
for(m=0;m<j+k;m++)i*=10;
a=i;
i=1;
for(m=0;m<k;m++)i*=10;
a-=i;
//cout<<b-d<<"/"<<a<<endl;
b=b-d;
k=b;j=a;
c=a;d=b;

while(y=j%k,y)
{
j=k;
k=y;
}


cout<<"Case "<<cases<<": "<<b/k<<"/"<<a/k<<endl;


}
}
[/cpp]

how can it be rte?
can anybody give some example?
thx

k + j <= 9 so the integer will work

if you use double you may suffer precision problem

Does it handle the case of j = 0 correctly? I still can't get it accepted, but this was one of the things I had to fix so far.

i don't know why i always get WA,can somebody help me ?thanks;
[cpp]
#include <iostream.h>
#include <math.h>
#include <cstdio>
#include <cstring>

int cas=1;

int gcd(int a,int b)
{
if (a%b==0)
return b;
else
return gcd(b,a%b);
}

int main()
{
char c[100];
int k,j;
int len;
double x;
double r;
int p,q;
int i;
while(cin>>j&&j!=-1)
{
cin.get();
cin.getline(c,100);
len=strlen(c);
k=len-j-2;
x=0;
r=0;
for(i=2;i<len;i++)
{
x+=int(c-48)*pow(10,-(i-1));
}
if(j!=0)
{
r=x*pow(10,k)-int(x*pow(10,k));
p=int(pow(10,k+j)*x+r-pow(10,k)*x);
q=int(pow(10,k+j))-int(pow(10,k));
cout<<"Case "<<cas++<<": "<<p/gcd(p,q)<<"/"<<q/gcd(p,q)<<endl;
}
else
{
p=int(x*pow(10,k));
q=int(pow(10,k));
cout<<"Case "<<cas++<<": "<<p/gcd(p,q)<<"/"<<q/gcd(p,q)<<endl;
}
}
return 0;
}


[/cpp]