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| Показать все ветки Спрятать все ветки Показать все сообщения Спрятать все сообщения | | to admin for input description | Ade | 1838. Удар самурая | 2 июн 2023 14:56 | 1 | "The second line contains n different positive integers, which are less than the L", That's result translated from RUS version.
But the ENG version has not the word "positive".
Please can you take a look? | | beautiful problem | Aneto | 1838. Удар самурая | 10 июл 2012 16:30 | 1 | just wanted to thank the author | | Samurai's Stroke | nvnp | 1838. Удар самурая | 13 май 2011 17:32 | 6 | Can the prop be at the end of the stick ?
Isn't it always length of stick for n=2 and n=3 ?
Did i understand the question correctly ?
This was giving WA.
v[0]=0;
int leftend=(v[1]-v[0])*2;
int rightstart=len- ((len-v[n])*2);
// 0 to leftend ,closest v[i];
// rightstart to len closest[j];
//
//
for(i=1;i<=n;i++){
if(leftend<=v[i]) { leftend=v[i];break; }
}
for(i=n;i>=1;i--){
if(rightstart>=v[i]) { rightstart=v[i];break; }
}
printf("%d\n",len-((rightstart-leftend)<0?0:((rightstart-leftend)))); every remaining peace should be remin at least at two props,
so, for n < 4, there no possible place to stroke (result = l)
Similar you can to ignore the part from 0 to v[2] and from v[n - 1] to l
prop can not be at the end of the stick, but it can be at strike-point (at end of remaining part)
You need to look all interval between v[i] and v[i + 1] for i in 2..(n-2) to find summ of strike-proof places, remaining part will fall, either outside (if strike too close to prop) or inside (if strike too far). So between every props there is only strike-proof interval.
Excuse my English :) Can you give an example ? I understand that Gennosuke cuts only once.
An example would help a lot.
I understood as follows
We have props at 1 2 3 4
0 1 2 3 4 5
[START]..v[1]..v[2].v[3]...[vn] XXXXXX[LEN]
now since the left overlang is v[1], for the center of mass to lie between 2 props the closest prop should >=2*v[1] and for the right overlang the center of mass to lie , it should be (len -v[n])*2 , so the closest prop <(len-v[n])*2 . We paint the overhang to these props. for example
10 4
2 3 7 8
we can't strike at left of 3, because left part will have only 1 prop and will fall down
like this we can't strike at right of 7
so, consider interval 3...7
at position less than 4 - left part of stick will fall down to left (because it's center of mass left-of first prop at position 2)
at position more than 6 - left part will fall to right,
also for right-part.
We can find the only interval we can use - it from 4 thru 6
So, we shall paint 10 - (6 - 4) = 8
Edited by author 01.05.2011 02:08 Can you give an example why we need to summuation of each interval and why we need to round it up ? I think it will always be an integer.
This gives me WA on test 7.
for(i=1;i<=n;i++) scanf(" %d",&arr[i]);
arr[i+1]=len;
/*
* v0,v1,v2,v3,v4,v5,v6
*/
int lovr = arr[1] ;
int rovr = len - arr[n] ;
int tot = 2*lovr + 2*rovr ;
int rmax=0,lmax;
for(i=n-1;i>=1;i--)
{
if( ( arr[n] - arr[i] ) > rovr )
{
if(i==n-1) {
rmax=len-arr[i];
}
else {
rmax=2*rovr;
}
break;
}
}
for(i=2;i<=n;i++)
{
if((arr[i]-arr[1]) > lovr)
{
if(i==2){
lmax=arr[i];
}
else
{
lmax=2*lovr;
}
break;
}
}
if(lmax + rmax > len) printf("%d\n",len);
else if(tot > len) printf("%d\n",len);
else printf("%d\n",rmax+lmax);
Edited by author 13.05.2011 16:36 try this
13 6
1 2 6 7 11 12
result 8 | | test | Ronnie | 1838. Удар самурая | 30 апр 2011 16:26 | 2 | test Ronnie 30 апр 2011 13:46 what's the answer for this?
5 2
0 5
if we cut anywhere, the centre of mass will be between two props, but i think it should fall.
so is it 0 or 5? The test is incorrect.
Because props can't stay on 0. Only >0. Sorry for bad english. | | WTF????? | Viktor | 1838. Удар самурая | 30 апр 2011 14:22 | 2 | WTF is this problem about o.O |
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