10023 - Square root

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Almost Human
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10023 - Square root

Post by Almost Human » Fri May 23, 2003 9:40 am

is there any special input to be considered ... ?

the input will always be an ordinary numbers right ?

Observer
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Post by Observer » Fri May 23, 2003 11:32 am

The input contains +ve integers only(is that what you mean by "ordinary numbers"?).

Bear in mind that this is a multiple-input question......

For sample i/o, I've only tried easy cases. For instance:
5

7206604678144

9152324687831194092834914521

186755313003091766118189319724929

1

0
Output:
2684512

95667782914789

13665844759951423

1

0
If you really couldn't figure out what's wrong, why don't you share your algorithm with us??
7th Contest of Newbies
Date: December 31st, 2011 (Saturday)
Time: 12:00 - 16:00 (UTC)
URL: http://uva.onlinejudge.org

Almost Human
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Post by Almost Human » Sat May 24, 2003 7:04 am

Code: Select all

#include <stdio.h>
#include <string.h>

int main ( void )
{
  unsigned long NumOfCase , length , i , j ;
  long double bedivide , divider ;
  char input[10000] ;

  freopen ( "10023.in" , "r" , stdin ) ;
  freopen ( "10023.out" , "w" , stdout ) ;

  scanf ( "%li\n\n" , &NumOfCase ) ;

  while ( NumOfCase -- )
  {
	 gets ( input ) ;
	 scanf ( "\n" ) ;

	 length = strlen ( input ) ;

	 if ( length % 2 )
	 {
		bedivide = input[0] - '0' ;
		i = 1 ;
	 }
	 else
	 {
		bedivide = ( input[0] - '0' ) * 10 - ( input[1] - '0' ) ;
		i = 2 ;
	 }

	 for ( divider = 0 ; ; )
	 {
		for ( j = 0 ; j <= 9 ; j ++ )
		  if ( ( divider * 10 + j ) * j > bedivide ) break ;

		j -- ;

		printf ( "%li" , j ) ;
		bedivide -= ( ( divider * 10 + j ) * j ) ;

		if ( input[i] == 0 ) break ;

		bedivide = bedivide * 100 + ( input[i] - '0' ) * 10 + ( input[i+1] - '0' ) ;
		i += 2 ;
		divider = divider * 10 + j + j ;
	 }
	 printf ( "\n\n" ) ;
  }

  return 0 ;
}
I've tried so many sample input range from 0 to 1e1000 it self but it's still WA...

please help

ec3_limz
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Location: Singapore

I got TLE

Post by ec3_limz » Mon May 26, 2003 1:00 pm

Sadly, I got a TLE for this problem. Any good algorithms out there?

[cpp]#include <stdio.h>
#include <string.h>

typedef struct bignum_t {
char d[2000];
};

int cmp(bignum_t a, bignum_t b) {
int i;

for (i = 0; i < 2000; i++) {
if (a.d < b.d)
return -1;
else if (a.d > b.d)
return 1;
}

return 0;
}

bignum_t sum(bignum_t a, bignum_t b) {
bignum_t rv;
int i;

memset(rv.d, 0, sizeof(rv.d));
for (i = 1999; i > 0; i--) {
rv.d += a.d + b.d;
rv.d += rv.d / 10;
rv.d %= 10;
}

return rv;
}

bignum_t getavg(bignum_t a, bignum_t b) {
bignum_t rv;
int i;

rv = sum(a, b);
for (i = 0; i < 2000; i++) {
if (i < 1999)
rv.d[i + 1] += (rv.d[i] % 2) * 10;
rv.d[i] /= 2;
}

return rv;
}

bignum_t multiply(bignum_t a, bignum_t b) {
bignum_t prod, sub;
int end1, end2, i, j, k;

for (end1 = 0; end1 < 2000 && a.d[end1] == 0; end1++);
for (end2 = 0; end2 < 2000 && b.d[end2] == 0; end2++);

memset(prod.d, 0, sizeof(prod.d));
for (i = 1999; i >= end1; i--) {
memset(sub.d, 0, sizeof(sub.d));

for (j = 1999, k = i; j >= end2; j--, k--) {
sub.d[k] += a.d[i] * b.d[j];
sub.d[k - 1] += sub.d[k] / 10;
sub.d[k] %= 10;
}

prod = sum(prod, sub);
}

return prod;
}

void process() {
char str[1010];
bignum_t target, lo, hi, avg, sq;
int comp, i, j;

scanf("%s", &str);
if (strcmp(str, "1") == 0) {
printf("1\n\n");
return;
}
memset(target.d, 0, sizeof(target.d));
memset(lo.d, 0, sizeof(lo.d));
memset(hi.d, 0, sizeof(hi.d));
for (i = strlen(str) - 1, j = 1999; i >= 0; i--, j--)
hi.d[j] = target.d[j] = str[i] - '0';

while (true) {
avg = getavg(lo, hi);
sq = multiply(avg, avg);

comp = cmp(sq, target);
if (comp == -1)
lo = avg;
else if (comp == 1)
hi = avg;
else
break;
}

for (i = 0; i < 2000 && avg.d[i] == 0; i++);
if (i >= 2000)
printf("0\n\n");
else {
while (i < 2000)
printf("%d", avg.d[i++]);
printf("\n\n");
}
}

int main() {
int ntest;

#ifndef ONLINE_JUDGE
freopen("input.dat", "r", stdin);
freopen("output.dat", "w", stdout);
#endif

scanf("%d", &ntest);
while (ntest-- > 0)
process();

return 0;
}
[/cpp]

Dominik Michniewski
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Post by Dominik Michniewski » Mon May 26, 2003 2:06 pm

I don't know is there still true, but I remember, than judge's input contians values Y such than X*X < Y and X is solution :(
Maybe it help us ...

Best regards
DM
If you really want to get Accepted, try to think about possible, and after that - about impossible ... and you'll get, what you want ....
Born from ashes - restarting counter of problems (800+ solved problems)

anupam
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Post by anupam » Tue May 27, 2003 6:39 am

shouldn't these should be removed if it is true?
--
i used string manipulation but can't solved.
i have tried all critical input i made.
but no prob.
so why wa.
--
anupam
"Everything should be made simple, but not always simpler"

anupam
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Post by anupam » Tue May 27, 2003 6:41 am

sorry, for mail and net problem same post was posted twice.
please forgive & forget.
anupam
"Everything should be made simple, but not always simpler"

Dominik Michniewski
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Post by Dominik Michniewski » Tue May 27, 2003 7:41 am

No problem.
I use string operations too and got TLE first time. I change only one thing to get Acc - I realize that my code may hang if and only if X*X < Y ... So I add one more condidtion and I got Acc .

DM
If you really want to get Accepted, try to think about possible, and after that - about impossible ... and you'll get, what you want ....
Born from ashes - restarting counter of problems (800+ solved problems)

jai166
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Posts: 10
Joined: Mon May 12, 2003 11:10 am

10023

Post by jai166 » Sat Jun 07, 2003 3:33 pm

Sorry, but why 10023 Accepted (P.E.)? Could you help me? Thank you!
[c]#include<stdio.h>
#include<math.h>
void main(void)
{
int i=0,t;
long double n[20];
while(scanf("%d",&t)==1){
printf("\n");
for(;i<t;i++){
scanf("%Lf",&n);
printf("\n");
}
for(i=0;i<t;i++)
printf("%.0Lf\n\n",sqrtl(n));
}
}[/c]

Whinii F.
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ooh..

Post by Whinii F. » Thu Jul 31, 2003 6:21 am

I suffered a few WAs from this problem and concluded that Dominik is definitely right. You should output largest X where X * X <= Y. The problem statement is completely wrong.

There should be a fix!
JongMan @ Yonsei

Joseph Kurniawan
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Location: Jakarta, Indonesia

Post by Joseph Kurniawan » Thu Jul 31, 2003 6:47 am

I think for this prob, you don't need special algo.
My friend told me to just use the command 'sqrtl'.
:wink:

jai166
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Posts: 10
Joined: Mon May 12, 2003 11:10 am

Thank you, and I've solved it!

Post by jai166 » Thu Jul 31, 2003 11:32 am

Thank you, and I've solved it! :D
[c]#include<stdio.h>
#include<math.h>
void main(void)
{
int t;
long double n;

scanf("%d",&t);
for(;t;t--){
scanf("%Lf",&n);
printf("%.0Lf\n",sqrtl(n));
if(t>1)putchar('\n');
}
}[/c]

minskcity
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Posts: 199
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Location: Vancouver

Post by minskcity » Sun Mar 21, 2004 6:54 am

Why does the judge accept these solutions? I thought Y can by 10^1000...????? :-? In which case this problem is not easy.

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sohel
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Location: New York

same question

Post by sohel » Sun Mar 21, 2004 7:16 am

Hmm,
I have the same question.
Can long double handle numbers of size 10^1000......... I don't think so.
Now then : why does the posted code get AC. :-?

20717TZ
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10023 I think this code is wrong, but accepted!!!!!!

Post by 20717TZ » Fri May 28, 2004 9:23 am

I think this code is wrong, but accepted!!!!!! Only use sqrtl()!!!!!!!!!!

[cpp]
//I think this code is wrong, but accepted!!!!!!
#include <cstdio>
#include <cmath>
void main()
{
FILE* fp=fopen("Input.txt","r");
int N,i;
long double Y;

fscanf(fp,"%d\n",&N);
for(i=0;i<N;i++)
{
fscanf(fp,"%Lf\n",&Y);
if(i)putchar('\n');
printf("%.Lf\n",sqrtl(Y));
}
}[/cpp]

Should sb. fix it
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