10494 - If We Were a Child Again

[Subeen]

I got WA in this problem. But can't find bugs in my programs.
Someone please give me some test input/output to test my program.
thanks...

[Hisoka]

input :

Code: Select all

234098720432984729 / 234923845
24324 / 2342

output :
check with calculator.

[Dominik Michniewski]

try also input which are longer than i.e. 20 characters and second input number is near to 2^31

Best regards
DM

[Observer]

"The first one (number) may be arbitrarily long."

How long will it approximately be? 100 chrs? Or more?

[Dominik Michniewski]

I assume that not longer that 2048 digits ....

Best regards
DM

[angga888]

1000 is enough.

Try for these input :

Code: Select all

0 / 2147483647 = 0
0 % 2147483647 = 0
2147483646 % 2147483647 = 2147483646

Good Luck


angga888

[Whinii F.]

Note that 32-bit integer is not enough to solve this problem. If B = 2^31 - 1, then the dividend could be too larger for a 32-bit integer to handle.

So, check this out:
99999999999999999999 / 2147483647

Result should be:
46566128752

[Subeen]

finally got accepted using long long int type variable.

thanks to all for their help

[sunhong]

I got several WA at this problem! But I can't find anything wrong in my program. And here is my code, please tell me what is incorrect. Thank you!

Code: Select all

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

int main()
 {long double a,b,result;
  char ch;
  while (cin>>a)
   {cin>>ch;
    while (ch!='/' && ch!='%') cin>>ch;
    cin>>b;
    if (ch=='/') result=a/b;
    else result=fmod(a,b);
    cout<<setiosflags(ios::fixed);
    cout<<setprecision(0)<<result<<endl;
   }
  return 1;
 }   

[Whinii F.]

Your program is surely wrong. -_-;

The input specifies that the first number to be arbitrarily large (Actually I think the test case contains numbers up to 1000 digits or so), but the long double variable's precision is not up to that.

You should implement bigint operation here. Read what is written so far.

[sunhong]

Thank you for your reply, but I'm not familiar with the bigint operation. So I don't know which datatype I should use to caculate the result, can you tell me please?

[Whinii F.]

Sure, why not.

You should use arrays to simulate the array as big integers. (when a[0] = 123 and a[1] = 456, these can be interpreted as 123456)
In case using strings as numbers are feasible, too.

Of course C or Pascal syntaxes do not support operations like these, so you should implement your own calculating functions like addition, subtraction, multiplication, .. etc.

There are algorithms to do operations of bigints faster than simulating hand operations (I believe there is one at Knuth's book), but in most cases, brute force operations are feasible.

For further info, try a search in this board using 'bigint' as the keyword.
Hope this helps.

[Observer]

Excuse me...

Can the result be stored in any data type (except BigInt)?


(I should think no........)

[Larry]

I didn't use a full BigInt class for this (because in a real contest, I wouldn't have brought one..)

But basically, you read in the first number as a string, the second number as an unsigned int, and then just output the thing in place.. it's pretty simple, since for %, it's obvious that the solution can fit in unsigned int, while you can use normal division rule..

[babor]

Hai guies I am fool at wa of 10494.
is my function wrong or where is wrong handling.
My code is
/*
/* @JUDGE_ID: XXXXXX 10494 C++ */
#include <stdio.h>
#include <string.h>
#define MAX 100000

unsigned long Modulas(char a[],unsigned long denomi)
{

unsigned long rim,len,i;
rim = (a[0] - 48 )%denomi;
len = strlen(a);
for(i=1;i<len;i++)
rim = ( 10*rim + (a - 48) )%denomi;
return rim;
}


char* String_divide(char a[],unsigned long denomi)
{
char res[MAX];
unsigned long rim,len,i,start,q=0;
rim =a[0]-48;
len = strlen(a);
for(i=1;i<len;i++)
{
rim = rim*10 + a-48;
if(rim>=denomi)
{
start = i;
break;
}
}

res[q++] = (rim/denomi) + 48;
rim = rim%denomi;
for(i=start+1;i<len;i++)
{
rim = ( 10*rim + (a - 48) );
res[q++] = (rim/denomi) + 48;
rim = rim%denomi;
}
res[q]=0;
return res;

}

void main()
{

char a[MAX],sign[3],result[MAX];
unsigned long rim,denomi;
while(scanf("%s %s %lu",a,sign,&denomi)==3)
{

if(strcmp(sign,"%")==0)
{

rim = Modulas(a,denomi);
printf("%lu\n",rim);
}
else
{
strcpy(result,String_divide(a,denomi));
printf("%s\n",result);
}

}

}
/*@END_OF_SOURCE_CODE*/




*/