OJ Board

Cleaned

Problems 474 and 545 are almost identical, but they (seem to) expect different output for n = 6 and n = 7 due to rounding variations:

474
2^-6 = 1.562e-2
2^-7 = 7.812e-3

545
2^-6 = 1.563E-2
2^-7 = 7.813E-3

Pls. tell me what's wrong with my code. I get WA but for 474 problem I got AC

I modified the code so it round the result now, that means that for n=6 I obtain 1.563E-2, and for n=7 => 7.813E-3...

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

main()
{ long n;
double u,a,l2;
long b;
l2=log10((double)2);
while(scanf("%ld",&n)==1)
{
printf("2^-%ld = ",n);
b=(long int)ceil((double)n*l2);
a=pow(2, b/l2-n);

u=((long int)(a*1000))/1000.0;
u=(a-u)*10000;
n=ceil(u);

a=(long int) (a*1000);
if (n>=5) a=a+1;
a=a/1000;
printf("%.3lfE-%ld\n",a,b);

}
}
[/cpp]

Thanx

Did you notice the blue flag? It's multi input!

Ok, but how must look this input?
and output


or

input:
and output

In fact I tried both but the same result WA...

try to use log2() instead of log10() .... log2() means log(), I think .....

I remember, that it was difference in rounding ....

Dominik Michniewski

I can't find log2() function in Borland C++ that I use. Ok no problem because I'll download gcc and I'll try again. Pls. tell me what about the input. Which of them are correct ?!

Thanx anyway

Both the input formats are correct. By the 2 in the first line you're saying that there will be two set of test cases. The input sets are separated by blank lines.

Who can email me a correct program?
Many thanks.
ccwangtao@sohu.com

for 545 and 474 , my outputs are same.
i think the only problem is in the inpur format of 545.

input is like that:

2

2
5

5
6

i don't use math theory.. ex) log..
but, my prog output is correct about other post's test cases..

i don't know whether my prog is worng or not.
is my prog really wrong??

or ain't i understanding the input format??
i saw that other post mentioned about the input format..

plz help me~

Not sure why I'm getting WA - is it an input format thing? (I know this is a slow way to do it).

[cpp]
#include <cstdio>

using namespace std;

int main() {
int n, c;
scanf("%d", &c);
for (int i = 0; i < c; ++i) {
if (i)
putc('\n', stdout);
scanf(" ");
int line = getc(stdin); ungetc(line, stdin);
while ((line != '\n') && (line != EOF)) {
scanf("%d", &n); while(getc(stdin) == ' ');
double mantissa = 1.0; int exp = 0;
for (int i = 0; i < n; ++i) {
mantissa /= 2.0;
if (mantissa < 1) {
mantissa *= 10.0;
++exp;
}
}
printf("2^-%d = %.3lfE-%d\n", n, mantissa, exp);
line = getc(stdin); ungetc(line, stdin);
}
}
}
[/cpp]

Your program gives different output from my AC solution:

diff me you
6,7c6,7
< 2^-6 = 1.563E-2
< 2^-7 = 7.813E-3
---
> 2^-6 = 1.562E-2
> 2^-7 = 7.812E-3

The correct numbers are

In[2]:=
\!\(2\^{\(-6\), \(-7\)} // N\)
Out[2]=
{0.015625, 0.0078125}

I have tried on all 9000 valid inputs, so that should fix it.

Isn't this issue addressed in an earlier post?

It gives the right values on my computer; I also tried it on several (linux) boxes in ECE. guess I'm just unlucky then.

ok, I'll just make those special cases.

I got tons of WAs before AC.

Algorithm is just the same as 474.

My friend got AC and we have the same answer for all n (1<=n<=9000) on windows XP + VC++6.0. But the judge always said WA.

Then I find the special precison lost when n = 6 and n = 7 which means

2^-6 = 1.56250E-2

2^-7 = 7.81250E-3

you have to face 5 when printf("%.3lf", answer)

Then I just printf("1.563") for n = 6 and printf("7.813") for n = 7 others are all same as the WA code and get AC.

The precison is really tedious problem.

The multy input is just like others

here is input and output

[cpp] scanf("%d", &cases);
gets(buf), gets(buf);
while (cases--)
{
while (gets(buf) && buf[0])
{
sscanf(buf, "%d", &n);
//calculate t and m
printf("2^-%d = %.3lfE%d\n", n, t, m);
}
if (cases)
printf("\n");
}[/cpp]

BTW i use log10 and pow functions
if you know why got WA on judge but same answer with AC code for all n on own windows plz tell me mail to:jackie@hit.edu.cn
THKS

Good luck