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| I am wrong ?? | xiaowentao99 | 1068. Sum | 27 Feb 2011 21:25 | 2 |
#include <iostream.h>
int main(){
int num;
while (cin>>num){
int answer=0;
if(num>=1){
answer = num*(num+1)/2;
printf("%d\n",answer);
}
else{
answer = (2-num)*(1+num)/2;
printf("%d\n",answer);
}
}
return 0;
}
where ? i faced same problem. compiler may not support cin>> but istream::cin>> or u should use
scanf("%d",&a); like this.
thanks |
| trouble(pascal) | Vyacheslav | 1079. Maximum | 27 Feb 2011 18:53 | 2 |
what`s wrong?
var n,i,res,c,max:longint;
a:array[0..100000] of integer;
begin
repeat
max:=0;
Readln(n);
a[0]:=0;
a[1]:=1;
for i:=1 to n do
begin
a[2*i]:=a[i];
a[2*i+1]:=a[i]+a[i+1]
end;
for i:=1 to n do
if a[i]>max then max:=a[i];
if n<>0 then Writeln(max);
until n=0;
end. i find mistake:
a:array[0..200001] of integer;
this is accepted) |
| WHY WA#16 | Nick Loginov (USU) | 1201. Which Day Is It? | 27 Feb 2011 17:16 | 2 |
WHY WA#16 Nick Loginov (USU) 26 Mar 2010 08:44 Try to check this: 25 11 1885.
The right answer is:
mon........2....9...16...23...30
tue........3...10...17...24
wed........4...11...18..[25]
thu........5...12...19...26
fri........6...13...20...27
sat........7...14...21...28
sun...1....8...15...22...29
Edited by author 27.02.2011 17:19 |
| Any idea where the mistake is? | Pawel Wanat | 1010. Discrete Function | 26 Feb 2011 19:10 | 5 |
This code gets WA on test 2:
[code deleted]
& this gets AC:
[code deleted]
:)
Edited by moderator 02.01.2020 18:26 Try to use
scanf("%I64d", &tab[i]); Oh! It's like in dev :)
thx for info. . The operating system may be windows, not linux, but why it return WA instead of CE?
Edited by author 27.07.2009 20:46 Cause your solution is wrong. |
| Is there any O(1) solution? | Sergey Zuev | 1017. Staircases | 26 Feb 2011 18:54 | 3 |
I know O(n^2) one. Maybe there is also an O(1)? I have an O(n*logn) one...but surely not good enough... |
| what's wrong? | Samirson | 1197. Lonesome Knight | 26 Feb 2011 16:14 | 1 |
#include <stdio.h>
int main(void) {
int n,k,i;
char c;
scanf("%d",&n);
while(n--){
scanf("%1s%d",&c,&k);
if((c=='a'||c=='h')&&(k==1||k==8)) {printf("2\n"); continue;}
if((c=='a'||c=='h')&&(k==2||k==7)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==1||k==8)) {printf("3\n"); continue;}
if((c=='b'||c=='g')&&(k==2||k==7)) {printf("4\n"); continue;}
if(c<'b'||c>'g') {printf("4\n"); continue;}
if(k<2||k>7) {printf("4\n"); continue;}
if((c=='b'||c=='g') || (k==2||k==7)) {printf("6\n"); continue;}
printf("8\n");
}
return 0;
} |
| What's wrong with my C# code? It is giving wrong answer in case # 2... | Alaska . PK | 1423. String Tale | 26 Feb 2011 00:03 | 1 |
long stringLength = Convert.ToInt32(Console.ReadLine());
string S = Console.ReadLine();
string T = Console.ReadLine();
long temp = 0;
bool check = false;
for (long i = 0; i < stringLength; i++)
{
S = S[Convert.ToInt32(stringLength) - 1] +
S.Substring(0, Convert.ToInt32(stringLength) - 1);
if (S == T)
{
temp = i + 1;
check = true;
break;
}
}
if (check && temp < stringLength)
Console.WriteLine(temp);
else
Console.WriteLine(-1);
|
| Note to the solution of the test number 3 | Alflex | 1294. Mars Satellites | 25 Feb 2011 21:49 | 1 |
The printing of the answer in C++ should be, e.g., in the form:
cout << "Distance is "
<< setiosflags(std::ios_base::fixed)
<< setprecision(0)
<< 1000.0*sqrt(dist) << " km.";
where dist is double value. You don't need to round off the answer with the help of some functions. |
| Test #1 for 1601 | Dejust | | 25 Feb 2011 17:46 | 1 |
Please give a test № 1 for the task № 1601 |
| To admins | Hakobyan Tigran (RAU) | 1005. Stone Pile | 25 Feb 2011 14:00 | 2 |
To admins Hakobyan Tigran (RAU) 9 Feb 2011 01:26 Why this solution got AC ? For the first test it outputs 15, but correct answer is 3 !!!! #include <iostream> #include <complex> #include <algorithm> #include <cmath> #include <vector> #include <string> using namespace std; int main() { int n,i,total=0,counter; int l; vector <int> x; cin>>n; for(i=0; i<n; i++) { cin>>l; x.push_back(l); } for(i=0; i<n; i++) { total+=x[i]; } counter=total/2; for(i=0; i<n; i++) { if(x[i]<=counter) { counter-=x[i]; } } cout<<2*counter+total%2; return 0; } |
| Test №4: Output limit exceeded | Vitaliy_Yurenya | 1313. Some Words about Sport | 24 Feb 2011 21:32 | 2 |
I have tested my program on many tests, and I don't see any mistakes in my algorithm. What does mean "Output limit exceeded"? Please help. Thanks. Hi. I got WA at test #4 unless I declared variables as Integer instead of 1..100. Maybe it'll help you, if you still have that error
Edited by author 25.02.2011 00:00 |
| So easy ! In this website, there are many problems similar to this problem. So, it costs me no efforts to solve it ! | Phan Hoài Nam (Harvey Nash) | 1282. Game Tree | 24 Feb 2011 15:14 | 1 |
Hint for you :
int travel(int player,int node)
{
int value;
if(isLeaf(node)) return getValue(node);
// If the first player is at this node, he prefer getting the maximal value among -1, 0 and +1.
if(player == 0)
{
value = MinValue();
for(int a=0;a<numberOfChild(node);a++)
{
int v = travel((player+1)%2,getChild(node,a));
if(value < v) value = v;
}
}
// Get the minimal value.
else
{
value = MaxValue();
for(int a=0;a<numberOfChild(node);a++)
{
int v = travel((player+1)%2,getChild(node,a));
if(value > v) value = v;
}
}
return value;
}
Edited by author 24.02.2011 15:14 |
| TLE 11 HELP PLZ !!!!!! | Nikitoz68 | 1024. Permutations | 24 Feb 2011 03:50 | 1 |
#include <cstdio>
using namespace std;
int main()
{
int const Q=1001;
long long unsigned k = 0,i,n,a=0;
scanf("%llu",&n);
long long unsigned mas1[Q];
long long unsigned mas2[Q];
for(i = 1;i<=n;i++)
{ scanf("%llu",&mas1[i]);
mas2[i] = mas1[i];
}
for(;;)
{
for(i = 1;i<=n;i++)
{
if (i == mas2[i]){ k++;}
mas2[i] = mas1[mas2[i]];
}
a++;
if ( k == n) { break;} else {k = 0;}
}
printf("%llu\n",a);
return 0;
} |
| WA №9 What's wrong? who can tell me plz? | Nikitoz68 | 1024. Permutations | 24 Feb 2011 02:56 | 1 |
#include <cstdio>
using namespace std;
/*
*
*/
int main()
{
long long k = 0;
long long i;
long long unsigned n;
scanf("%llu",&n);
long long unsigned mas1[1000];
long long unsigned mas2[1000];
for(i = 1;i<=n;i+=1)
{ scanf("%llu",&mas1[i]);
mas2[i] = mas1[i];
}
long long unsigned a = 0;
long long s = 0;
long long m = 0;
do
{
for(long long j = 1;j<=n;j+=1)
{
if (j == mas2[j]){ k+=1;}
}
for(i = 1;i<=n;i+=1)
{
s = mas2[i];
mas2[i] = mas1[mas2[i]];
}
if ( k == n) { m=1;} else {k = 0;}
a+=1;
} while (m!=1);
printf("%llu\n",a);
return 0;
} |
| Warning: newline is required | Evgeniy++ | 1079. Maximum | 24 Feb 2011 00:54 | 2 |
Hello everyone!
My program was accepted only with 'newline' output for each test case. Before that I`ve got I/A.
You have been warned :)
See you! |
| Wrong Answer. What's wrong in my program? I'm sure that i'm right! №1024 | Nikitoz68 | | 24 Feb 2011 00:51 | 1 |
#include <cstdio>
using namespace std;
/*
*
*/
int main()
{
unsigned int k = 0;
unsigned int i;
unsigned int n;
scanf("%u",&n);
unsigned int mas1[1000],mas2[1000];
for(i = 1;i<=n;i+=1)
{ scanf("%u",&mas1[i]);
mas2[i] = mas1[i];
}
unsigned int a = 0;
unsigned int s = 0;
unsigned int m = 0;
do
{
for(int j = 1;j<=n;j+=1)
{
if (j == mas2[j]){ k+=1;}
}
for(i = 1;i<=n;i+=1)
{
s = mas2[i];
mas2[i] = mas1[s];
}
if ( k == n) { m=1;} else {k = 0;}
a+=1;
} while (m!=1);
printf("%u\n",a);
return 0;
} |
| WA 27 | Adrian Zaharia | 1238. Folding | 23 Feb 2011 23:36 | 1 |
WA 27 Adrian Zaharia 23 Feb 2011 23:36 Can I have a test for WA 27 please? |
| Getting Access Violation on Test #1 (Problem 1471) | Kinan Sarmini | | 23 Feb 2011 23:10 | 1 |
It's working now, had a simple bug.
Edited by author 23.02.2011 23:34 |
| WA10 | Серовиков Андрей | 1564. Counting Ones | 23 Feb 2011 21:32 | 8 |
WA10 Серовиков Андрей 13 Oct 2007 14:53 What's wrong, if the second test is 10^18?
the ones of 1 to 10^18 have over the range of int64 Re: WA10 Piratek-(akaDK) 16 Mar 2008 18:14 The number of ones is less than number of floor. For me tha answer must be in Int64 Re: WA10 Vedernikoff Sergey 17 Mar 2008 22:03 The answer is within Int64. So, try to find bug in your code or algorithm... Re: WA10 Chmel_Tolstiy 18 Mar 2008 19:44 if you use line like this at TIMUS will a bug ...
long long INF = 1e18;
you should use
long long INF = 1e9;
INF *= INF; Use unsigned __int64, might help. Also, if you use binary search, make sure that left+right does not overflow prior to /2 or >>1. if you use line like this at TIMUS will a bug ...
long long INF = 1e18;
you should use
long long INF = 1e9;
INF *= INF; Thank you! Try this test
12
Answer 19 (not 20) |
| Is the test case correct??? | 72VanVector[SevNTU] | 1817. Merry-go-round | 22 Feb 2011 05:58 | 15 |
consider the test case in problem statement. Will the kid really wait for 0.6875 sec, when 2 kids are riding merry-go round?? Mb this value should be 0.625 = 5/8???
Edited by author 30.01.2011 13:22 The test case is correct. it seems to be wrong at all
there are 6 posibilities in this hypothesis (there are 2 childrens) they are:
1100 - 2
1010 - 1
1001 - 1
0110 - 0
0101 - 0
0011 - 0
so, the posibility must be 2*1/6 + 1*1/6 + 1*1/6 = 0.66666667
isn't it right??
Edited by author 30.01.2011 14:15
_________________________________________
test case is right. there are sub-hypothesis with different posibilities
Edited by author 30.01.2011 14:54 Seems like the possibility of each case isn't equal to 1/6, some of the cases have more then 1/6, others - less. How did you do that??
Edited by author 30.01.2011 16:45 Yes, test case is correct.
Look:
1010 you can get from 1000 and 0010, p = 2/16
0101 you can get from 0100 and 0001, p = 2/16
But:
1100 you can get from 1000 (2 times!) and 0100, p = 3/16
0110 you can get from 0100 (2 times!) and 0010, p = 3/16
0011 you can get from 0010 (2 times!) and 0001, p = 3/16
1001 you can get from 0001 (2 times!) and 1000, p = 3/16
Thus, we get:
p(1100) = 3/16, p(1001) = 3/16, p(1010) = 2/16
And the answer is:
(3/16)*2 + (3/16)*1 + (2/16)*1 = 11/16 = 0.6875
Nice problem! Look:
1010 you can get from 1000 and 0010, p = 2/16
0101 you can get from 0100 and 0001, p = 2/16
But:
1100 you can get from 1000 (2 times!) and 0100, p = 3/16
0110 you can get from 0100 (2 times!) and 0010, p = 3/16
0011 you can get from 0010 (2 times!) and 0001, p = 3/16
1001 you can get from 0001 (2 times!) and 1000, p = 3/16 Can you please explain why for 1010 -> p = 2/16 but 1100 -> p = 3/16 How does 1000 come 2 times in this case ? 1000 -> 1010 if start position = 3
0010 -> 1010 if start position = 1
(2 times)
1000 -> 1100 if start position = 1
1000 -> 1100 if start position = 2
0100 -> 1100 if start position = 1
(3 times) "And the answer is:
... + (3/16)*1 + = 11/16 = 0.6875"
Could you explain why (3/16)*1 but not (3/16)*2,
though max waiting time is 2 (1001 - if we stand at place 1)??
Mask Time
0011 0
0101 0
0110 0
1001 1
1010 1
1100 2 Sorry, but why 11/16? Why not 3/16 + 3/16 + 3/16 + 3/16 + 2/16 + 2/16 = 16/16
I understand that
4/16 == 0.250000 ======
1000 1/16
0100 1/16
0010 1/16
0001 1/16
24/16 == 1.500000 =====
1110 (3+2+1=6)/16
1101 (3+2+1=6)/16
1011 (3+2+1=6)/16
0111 (3+2+1=6)/16
But why 11/16? Just some basic knowledge from probability theory...
3/16 + 3/16 + 3/16 + 3/16 + 2/16 + 2/16 = 16/16
It's a normalization requirement (p1 + p2 + ... + pn = 1).
And mean MX = p1*x1 + p2*x2 + ... + pn*xn,
where pi - probabilities, xi - values.
So we have (3/16)*2 + (3/16)*1 + (2/16)*1 + (3/16)*0 + (3/16)*0 + (2/16)*0 = 11/16. In Buffon problem there are 3 different
solution and all of them are right.
Answer depend on exact understanding
all conditions of random experiment.
I think that this problem also safer from too
brief explanation.
i've used this explanation of test:
let it be some kind of class:
1100 0110 0011 1001 - first class, maked by all rotations of first one, and it's representative would be a smallest, so:
[1100] : {1100 0110 0011 1001}
[1010] : {1010 0101 1010 0101} - second class, we have repeatings here, but it must be always n rotations
so, we could get [1100] from:
1000 - waiting 1 sec,
0001 - not waiting
0100 - not waiting
all of them is class [1000] which comes from [0000] with 100% posibility
so posibilities of [1000] would be:
1000 - 100% * 1 / 4
0001 - 100% * 1 / 4
0100 - 100% * 1 / 4
so posibility of getting in class [1100] is 3/4
[1010] we could get from:
0010 which posibility is 100% * 1 / 4
so our classes have such posibilities:
[1100] - 3 / 4
[1010] - 1 / 4
[1100]:
mask sec posibility
1100 2 1/4*3/4
0110 0 1/4*3/4
0011 0 1/4*3/4
1001 1 1/4*3/4
[1010]:
mask sec posibility
1010 1 1/4*1/4
0101 0 1/4*1/4
1010 1 1/4*1/4 - it repeats, but it is another rotation, 3rd
0101 0 1/4*1/4 - 4th
so, our result would be:
MX = 2 * 3/16 + 1 * 3/16 + 1 * 1/16 + 1 * 1/16 = 11/16 = 0.6875
like in our test case
i've used DP for this problem, just filling array of posibilities in recursive with saving way, rotating my current mask, and looking from where i could go here |
