276 - Egyptian Multiplication

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ganza
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Post by ganza » Wed Oct 17, 2001 1:35 am

Hi!

I've sent that problem to judge and got WA

What should be done when a multiplication of a pair exceeds 99999?

Thanks,
Nuno Teixeira

PS: This board is great!! No more popups!!

FCS
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Post by FCS » Wed Oct 17, 2001 4:18 am

Do all calculations modulo 100000.

ganza
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Post by ganza » Thu Oct 18, 2001 4:55 pm

Now that works fine!

Thanks a lot!

FlyDeath
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Post by FlyDeath » Thu Jan 03, 2002 4:44 pm

What do you means??I don't understand
If i enter "r n n" then what shoud it be??
Here is the answer my program gave:

| r
|| * rr
|||| rrrr
|||||||| * rrrrrrrr
The solution is:
(I didn't type the space properly above)

ganza
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Post by ganza » Fri Jan 04, 2002 2:09 am

yep!
My program gets the same result

Do you get WA(Wrong Answer)?

FlyDeath
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Post by FlyDeath » Fri Jan 04, 2002 2:05 pm

I got AC finally

keen
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276 - what

Post by keen » Sun May 18, 2003 6:29 pm

Please, can anybody tell me why I get WA?
[c]
#include <stdio.h>
#include <string.h>

int egiDec (char num[55]);
void imprimeEgi (int x, int ast);

int main()
{
int esquerda=0, i=0, j=0, k=0, m=0, decimal=0, decimal2=0, resultado=0;
char num[55] = "";

fgets (num, 55, stdin);
while (num[0] != '\n')
{
decimal = egiDec (num);
fgets (num, 55, stdin);
decimal2 = egiDec (num);
esquerda = 1;
resultado = (decimal * decimal2) % 100000;
while (esquerda * 2 <= decimal2)
esquerda *= 2;
i = decimal2;
for (j = 1; j <= esquerda; j *= 2)
{

k = esquerda;
m = k;
while (m < i)
{
k /= 2;
if (m + k <= i)
m += k;
}
if (k == j)
{
imprimeEgi (j, 1);
i -= j;
}
else
imprimeEgi (j, 2);
imprimeEgi (decimal, 0);
decimal *= 2;
}
printf ("The solution is: ");
imprimeEgi (resultado, 0);
fgets (num, 55, stdin);
}
return 0;
}

int egiDec (char num[55])
{
int lenNum=strlen(num), i=0, j=0;
int dec=0;
char grupo[15]="";

for (i=0; i<=lenNum-1; i++)
{
if (num == ' ')
{
grupo[j] = '\0';
switch (num[i-1])
{
case '|' :
dec += strlen(grupo);
break;
case 'n' :
dec += 10 * strlen(grupo);
break;
case '9' :
dec += 100 * strlen(grupo);
break;
case '8' :
dec += 1000 * strlen(grupo);
break;
case 'r' :
dec += 10000 * strlen(grupo);
break;
}
j = 0;
}
else if (num != '\0')
{
grupo[j] = num;
j++;
}
else
break;
}
return dec;
}

void imprimeEgi (int x, int ast)
{
int casas[5] = {0, 0, 0, 0, 0};
int i=0, len=0;
char a;
for (i = 0; i <= 4; i++)
{
casas = x % 10;
x /= 10;
}
for (i = 0; i <= 4; i++)
{
switch (i)
{
case 0:
a = '|';
break;
case 1:
a = 'n';
break;
case 2:
a = '9';
break;
case 3:
a = '8';
break;
case 4:
a = 'r';
}
for (; casas > 0; casas--)
{
printf ("%c", a);
len++;
}
if ((len > 0) && (i < 4) && (casas[i+1] > 0))
{
printf (" ");
len++;
}
}
printf (" ");
len++;
if (ast == 1)
{
printf ("*");
len++;
}
if (ast != 0)
{
for (; len < 34; len++)
printf (" ");
}
else
printf ("\n");
}
[/c]

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angga888
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Location: Indonesia

276 - Egyptian Multiplication WA

Post by angga888 » Sun Jun 22, 2003 1:58 pm

Hello, help me.
I'm getting WA over and over. :cry:
Please give me the output for these cases :

Code: Select all

r
n
| n 9 8 r
rrrrrrrrr
rrrrrrrrr
rrrrrrrrr
||||||||| nnnnnnnnn 999999999 888888888 rrrrrrrrr
||||||||| nnnnnnnnn 999999999 888888888 rrrrrrrrr
And I don't quite understand, what to do if the number while in steps (or the result) getting overflow (>99999). I just do modulus 100000 everytime I get new number.
Are there any blank lines? or any tricky inputs?

Thanks. :D

Regards,
angga888

Per
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Location: Sweden

Post by Per » Sun Jun 22, 2003 2:56 pm

My solution outputs:

Code: Select all

|                                 r
|| *                              rr
||||                              rrrr
|||||||| *                        rrrrrrrr
The solution is:
|                                 | n 9 8 r
||                                || nn 99 88 rr
||||                              |||| nnnn 9999 8888 rrrr
||||||||                          |||||||| nnnnnnnn 99999999 88888888 rrrrrrrr
|||||| n *                        |||||| nnnnnnn 9999999 8888888 rrrrrrr
|| nnn                            || nnnnn 99999 88888 rrrrr
|||| nnnnnn                       |||| 9 8 r
|||||||| nn 9 *                   |||||||| 99 88 rr
|||||| nnnnn 99 *                 |||||| n 9999 8888 rrrr
|| n 99999 *                      || nnn 99999999 88888888 rrrrrrrr
|||| nn 8 *                       |||| nnnnnn 999999 8888888 rrrrrrr
|||||||| nnnn 88 *                |||||||| nn 999 88888 rrrrr
|||||| nnnnnnnnn 8888 *           |||||| nnnnn 999999 r
|| nnnnnnnnn 9 88888888           || n 999 8 rr
|||| nnnnnnnn 999 888888 r *      |||| nn 999999 88 rrrr
|||||||| nnnnnn 9999999 88 rrr    |||||||| nnnn 99 88888 rrrrrrrr
|||||| nnn 99999 88888 rrrrrr *   |||||| nnnnnnnnn 9999 rrrrrrr
The solution is: rrrrrrrrr
|                                 rrrrrrrrr
||                                rrrrrrrr
||||                              rrrrrr
||||||||                          rr
|||||| n *                        rrrr
|| nnn                            rrrrrrrr
|||| nnnnnn                       rrrrrr
|||||||| nn 9 *                   rr
|||||| nnnnn 99 *                 rrrr
|| n 99999 *                      rrrrrrrr
|||| nn 8 *                       rrrrrr
|||||||| nnnn 88 *                rr
|||||| nnnnnnnnn 8888 *           rrrr
|| nnnnnnnnn 9 88888888           rrrrrrrr
|||| nnnnnnnn 999 888888 r *      rrrrrr
|||||||| nnnnnn 9999999 88 rrr    |||| 9999999 88 rrrrr
|||||| nnn 99999 88888 rrrrrr *   |||| 9999999 88 rrrrrrr
The solution is: |||||||| 9999 88888 rrrrrr
| *                               ||||||||| nnnnnnnnn 999999999 888888888 rrrrrrrrr
|| *                              |||||||| nnnnnnnnn 999999999 888888888 rrrrrrrrr
|||| *                            |||||| nnnnnnnnn 999999999 888888888 rrrrrrrrr
|||||||| *                        || nnnnnnnnn 999999999 888888888 rrrrrrrrr
|||||| n *                        |||| nnnnnnnn 999999999 888888888 rrrrrrrrr
|| nnn                            |||||||| nnnnnn 999999999 888888888 rrrrrrrrr
|||| nnnnnn                       |||||| nnn 999999999 888888888 rrrrrrrrr
|||||||| nn 9 *                   || nnnnnnn 99999999 888888888 rrrrrrrrr
|||||| nnnnn 99                   |||| nnnn 9999999 888888888 rrrrrrrrr
|| n 99999 *                      |||||||| nnnnnnnn 9999 888888888 rrrrrrrrr
|||| nn 8 *                       |||||| nnnnnnn 999999999 88888888 rrrrrrrrr
|||||||| nnnn 88                  || nnnnn 999999999 8888888 rrrrrrrrr
|||||| nnnnnnnnn 8888             |||| 999999999 88888 rrrrrrrrr
|| nnnnnnnnn 9 88888888           |||||||| 99999999 8 rrrrrrrrr
|||| nnnnnnnn 999 888888 r        |||||| n 999999 888 rrrrrrrr
|||||||| nnnnnn 9999999 88 rrr *  |||||| nnn 999999999 888888888 rrrrrrrrr
|||||| nnn 99999 88888 rrrrrr *   || nnnnnnn 99999999 888888888 rrrrrrrrr
The solution is: ||||||||| 9999 88888 rrrrrr
As you can see, I only do the modulus when presenting the end result. (It sounds rather weird, but it's been a while since I solved the problem, and well, it's accepted :))

As far as I can tell, my solution would terminate on a blank line, so I don't think there are any.

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angga888
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Location: Indonesia

Post by angga888 » Wed Jun 25, 2003 9:29 am

Thanks Per for your output. :)

But I want to ask for input

Code: Select all

rrrrrrrrr
rrrrrrrrr
means 90000 * 90000 right?
So the solution if mod by 100000 should be 0 :wink:
Your output is |||||||| 9999 88888 rrrrrr
Can you explain to me why?

What variable type do you use? int, long, or long long?
And in what way do you calculate the result?
By adding the right number when there is an asterisk on left number,
finally mod 100000, or
Just calculate (a*b) mod 100000 and ignoring the process at all?

You said that you don't do modulus at all while in process, so
if the number (left or right) getting bigger (overflow from max variable
you use), what will happen?


Thanks :D


Regards,
angga888

Per
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Location: Sweden

Post by Per » Wed Jun 25, 2003 11:57 am

I use int, and just print the final result as (a*b)%100000, so when the input is 90000*90000 I get an overflow (and that's why my answer was not 0) :oops:

In such cases I would get overflow in the process as well.

So I'm guessing there are no test cases in judge's input where a*b >= 2^31

Larry
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Post by Larry » Mon Aug 25, 2003 7:58 am

Can someone post more inputs? I keep getting WA... :-?

Larry
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Post by Larry » Mon Aug 25, 2003 8:21 am

Nevermind, I tried multiply by 0 when input is a blank line, and I changed it to return when that happens and got AC. Stupidness.

anupam
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Post by anupam » Mon Aug 25, 2003 8:49 am

well, what is the rule of the spaces.
i had wa for the spaces.
please make it clear and precise.
--
Anupam :oops:
"Everything should be made simple, but not always simpler"

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mlvahe
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Location: Yerevan, Armenia

Help me please.

Post by mlvahe » Wed Oct 27, 2004 6:53 pm

:cry: :cry: :cry: :cry: :cry:
Please help me. Because whatever I do I get WA.
I don't know what to do.
[cpp]
#include <iostream>
using namespace std;

char dig[] = {'|','n','9','8','r'};

int print(int a)
{
int i = 0, l = 0, j;
while (a != 0)
{
if (a % 10)
{
for (j = 0; j < a % 10; j++)
{
cout << dig;
l++;
}
cout << ' ';
l++;
}
i++;
a /= 10;
}
return l;
}

int getdeg(char c)
{
int i, t10 = 1;
for (i = 0; i < 5; i++)
{
if (dig == c)
return t10;
t10 *= 10;
}
return 0;
}

int translate(char *a)
{
int i, s = 0;
for (i = 0; a; i++)
s += getdeg(a);
return s;
}

main()
{
char str[1000];
int a, b, i, l, a0, j, m;
while (cin.getline(str,1000))
{
b = translate(str);
if (!(*str))
return 0;
cin.getline(str,1000);
a = translate(str);
/* if (!(*str))
return 0;*/
a0 = a;
m = 1;
i = 1;
while (a)
{
l = print(m);
if (a % 2)
{
cout << '*';
l++;
}
for (j = l + 1; j < 35; j++)
cout << ' ';
print(m*b);
m *= 2;
a /= 2;
i++;
cout << '\n';
}
cout << "The solution is: ";
print((a0*b)%100000);
cout << '\n';
}
return 0;
}
[/cpp]

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