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| number of test cases ? | m313 | | 8 Dec 2014 21:25 | 1 |
hi
what is the total number of test cases for each problem ? |
| Stuck at WA#3 | begi | 1098. Questions | 8 Dec 2014 20:13 | 2 |
Can someone provide more tests?
I am using scanner.nextLine() to read
line.charAt(pos) == '?' to compare in java and also Josephus allgorithm:
jos[1] = 1;
for (int i = 2; i < max_n; i++) {
jos[i] = (jos[i - 1] + 1998) % i + 1;
}
My code is so small that I don't even know where is the bug Ok, the thing is I misunderstood the problem.
You read ALL the input ignoring new lines, and then do the 'process'.
I was reading line by line and doing 'process' for each line.
Hope it will help the others. |
| what is test N14? | Rati | 1276. Train | 7 Dec 2014 21:52 | 1 |
|
| 13th test | Popovici Mircea | 2011. Long Statement | 7 Dec 2014 15:16 | 8 |
I'm wondering if anyone knows the 13th test... i think it's some kind of special case. These are 3 test i suppose cause, it doesn't specificate that we have multiple test cases in one input. Am I right? No, you are not right. There is just one test case in every input, multiple tests were written just for you.
Test:
7
1 1 1 1 1 1 2
YES Test:
4
1 1 2 2
Ans:
Yes
Edited by author 11.11.2013 15:19 4
1 1 2 2
Yes
cause:
1122
1212
1221
2121
2211
2112
Edited by author 16.12.2013 23:57 i thinks these tests are not suitable for 13th s. 'casue i passed these ,but get 13th wrong. Two-Eight-Nine.Thanks...!!! |
| Wrong Answer C ! Help please! | Pedro | 1001. Reverse Root | 6 Dec 2014 21:43 | 2 |
#include <stdio.h>
#include <math.h>
int main(){
double num1,num2,num3,num4;
scanf("%lf %lf %lf %lf",&num1,&num2,&num3,&num4);
printf("%.4lf\n%.4lf\n%.4lf\n%.4lf",sqrt(num4),sqrt(num3),sqrt(num2),sqrt(num1));
return 0;
}
why is it wrong? I think it's not necessarily exactly 4 numbers.
Judging from my result, I think there are many numbers in the test case.
So you need a list or something to store the integers first then print the sqrt in reversed order.
(also, the integer is very big, so int type might not work) |
| wa3 | Ade [FDU] | 1317. Hail | 6 Dec 2014 16:07 | 1 |
wa3 Ade [FDU] 6 Dec 2014 16:07 check the cases when you shoot hail over fence's westmost edge. |
| acos(min(1, angle)) | Ade [FDU] | 1331. Vladislava | 6 Dec 2014 13:46 | 1 |
|
| good test for WA#4 | hoan | 1247. Check a Sequence | 5 Dec 2014 22:33 | 2 |
input:
6 10
2
2
2
0
10
1
output:
NO
|
| for wa5 | Ade [FDU] | 1130. Nikifor's Walk | 5 Dec 2014 21:05 | 1 |
9
9
0 9
9 0
5 4
4 5
0 9
9 0
-5 4
-4 5
9 0 |
| Why i got WA?Can anybody tell me? Thanks! | TaoZhang | 1131. Copying | 5 Dec 2014 20:33 | 3 |
Here is my program
var
n,k,i,time:longint;
begin
readln(n,k);
if k=1 then writeln(n-1)
else
begin
i:=0;time:=0;n:=n-1;
while n>0 do
begin
if i<k then i:=i+1;
n:=n-i;time:=time+1;
end;
writeln(time);
end;
end. > Here is my program
> var
> n,k,i,time:longint;
>
> begin
> readln(n,k);
> if k=1 then writeln(n-1)
> else
> begin
> i:=0;time:=0;n:=n-1;
> while n>0 do
> begin
>(***) if i<k then i:=i+1;
> n:=n-i;time:=time+1;
> end;
> writeln(time);
> end;
> end.
Have a look at the iteraton..as I understand i is the number of used
cables... But on each iteration you should not add only one cable
(***) but as much as possible, I mean if you have used i cables by
now, next time you could use 2*i cables, if you have so much, NOT
i:=i+1!
Sample input:
12 6
Sample output:
4
1 sec. 1-2
2 sec. 1-3, 2-4
3 sec. 1-5,2-6,3-7,4-8
4 sec. 1-9,2-10,3-11,4-12
Bye!
and Have a luck :) |
| Easy | Nodir NAZAROV Komilijonovich [TUIT-Karshi] | 2031. Overturned Numbers | 5 Dec 2014 19:28 | 1 |
Easy Nodir NAZAROV Komilijonovich [TUIT-Karshi] 5 Dec 2014 19:28 only 4 cases
Edited by author 05.12.2014 19:38
Edited by author 05.12.2014 19:39 |
| Mistake | Nodir NAZAROV Komilijonovich [TUIT-Karshi] | 2035. Another Dress Rehearsal | 5 Dec 2014 18:55 | 1 |
Mistake Nodir NAZAROV Komilijonovich [TUIT-Karshi] 5 Dec 2014 18:55 found my mistake. sorry for this post.
Edited by author 05.12.2014 22:16
Edited by author 05.12.2014 22:16 |
| сколько всего тестов в этой задаче? | СУНЦ УрФУ petrov nikita | 1415. Mobile Life | 5 Dec 2014 13:37 | 2 |
You will not solve this problem (((
Edited by author 05.12.2014 13:37
Edited by author 05.12.2014 13:37 |
| if somebody had an 2 and 10-11 WA | daemon | 1100. Final Standings | 5 Dec 2014 12:59 | 3 |
if 2 - try
2
4 0
3 2
if 10-11 - just not use sort. especially bubble one))) I have 2 and this test works correct
Edited by author 05.12.2014 13:00 |
| WA14 | Makar | 1419. Maps of the Island Worlds | 4 Dec 2014 21:47 | 1 |
WA14 Makar 4 Dec 2014 21:47 Can't get it. What in this test case.
P.S. finding bridges from e-maxx with deep of recursion. |
| very annoying problem | arrammis | 2011. Long Statement | 4 Dec 2014 20:48 | 1 |
i had 9 wroung attempts, 10 was AC lol |
| WA testcase #1 | Manish Jangir | 1837. Isenbaev's Number | 4 Dec 2014 18:18 | 2 |
i tried BFS. but WA in test #1. Can anyone suggest some test cases?
Thanks.. Don't output last CRLF. Bad cheker. |
| What about 4-th test?Please help.Here is source: | Search | 1617. Flat Spots | 4 Dec 2014 17:01 | 10 |
#include<iostream> #include<algorithm> using namespace std; int main() { int i,n,a[155],k=0; cin>>n; for(i=0;i<n;i++) cin>>a[i]; sort(a,a+n+1); for(i=0;i<n;i++) if(a[i]==a[i+3]) k++; cout<<k; return 0; } Re: * Search 1 Apr 2008 20:04 Re: * tolia 7 Mar 2009 18:40 #include<iostream> #include<algorithm> using namespace std; int main() { int i,n,a[155],k=0; cin>>n; for(i=0;i<n;i++) cin>>a[i]; sort(a,a+n+1); for(i=0;i<n;i++) if(a[i]==a[i+3]) k++; cout<<k; return 0; } Re: * Adhambek 26 May 2013 09:48 look at my test .... use your program
10
600
600
600
600
600
600
600
600
600
600
I think your answer is 7... But true answer is 2!
Edited by author 26.05.2013 09:50 Re: * Ivan Metelev 4 Dec 2014 17:01 Sorry, but true answer is 1, on my opinion
Edited by author 04.12.2014 17:01
Edited by author 04.12.2014 17:02 Re: * LLIRIK 3 Apr 2008 22:59 5
600
600
600
600
600 what is the answer for this test? my programm writes 1 isn't it right?? Re: * Rashid Gaziyev 26 Oct 2010 11:00 |
| My approach-will it be correct | Ankit Tripathi | 1017. Staircases | 4 Dec 2014 10:45 | 1 |
I construct vertical blocks made of bricks. Now, blocks are of height 1,2,3,...n-1. Now, to make these many i need n*(n-1)/2 bricks. But, I have n bricks so I have to remove former minus latter number of bricks, which is n(n-3)/2. Now, the problem can be changed to number of ways of obtaining n(n-3)/2 from numbers 1,2,3,...n-1 while each number can be used only once. If you find the approach to be correct, please tell me how to obtain this number.(I am a newbie in coding..:)). |
| Incorrect reasoning ? | artem | 1225. Flags | 4 Dec 2014 02:12 | 2 |
Let me know please, why i should not think in this way -
i have A ways to make n-1 strip sequences. If i will add red or white strip from the
left side, i have A ways to make n strip sequences.
R/W + [n-1.....]
But in the same time i can add red or white strip from the right side, and so i'll have another
A ways to make n strip sequences.
[n-1....] + R/W
So, now i have 2*A ways to make n strip sequences.
What's wrong in this kind of argumentation ? Some ways may be counted twice, so the ways will be less than 2A |
