input ai bi C
it means that ->   line [ai, bi) is colored by C.

Can anyone please provide test case #11? I am getting WA on that.

Could I get the data of the test 3

i've been thinking at this problem for 2 weeks. can somebody give me a hint ?

My code is :
#include<iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct slot{
    int start;
    int stop;
};

bool order1(const slot &lhs,const slot & rhs)
{
    return lhs.stop<rhs.stop;
}

bool order2(const slot &lhs,const slot & rhs)
{
    return lhs.start<rhs.start;
}

int main()
{
    int timemax,timemin;
    long n,i,j,temp,dp[30002],max;
    slot a[100001];
    dp[0]=0;
    scanf("%ld",&n);
    for(i=0;i<n;i++)
        scanf("%d%d",&a[i].start,&a[i].stop);

    sort(a,a+n,order2);
    timemin=a[0].start;
    stable_sort(a,a+n,order1);
    timemax=a[n-1].stop;

    for(i=0;i<=timemin;i++)
        dp[i]=0;
    for(i=timemin+1;i<=timemax;i++)
    {
        max=dp[i-1];
        while(a[j].stop<i && j<n)
        j++;
        while(a[j].stop==i)
        {
            if(a[j].start-1>=0)
                temp=1+dp[a[j].start-1];
            else
                temp=1;
            if(temp>max)
                max=temp;
            j++;
            if(j==n)
                break;
        }
        dp[i]=max;
    }



    printf("%ld\n",dp[timemax]);
}



I dont know why I am getting wrong answer on test case 1 and runtime error on Ideone.com

Условия:
> найти для экскурсии кратчайший маршрут, начинающийся и заканчивающийся в одном и том же месте.
> Ваша программа должна найти маршрут наименьшей длины
> и M двусторонних дорог,

В ТЗ не сказано что нужно обойти максимум перекрестков, не сказано что нельзя ехать назад.
А значит путь: 1 -> 4 -> 1 подходит.

Wrong #2 please help !!!

hi
what is the total number of test cases for each problem ?

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

I'm wondering if anyone knows the 13th test... i think it's some kind of special case.

#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?

check the cases when you shoot hail over fence's westmost edge.

input:
6 10
2
2
2
0
10
1

output:
NO

9
9
0 9
9 0
5 4
4 5
0 9
9 0
-5 4
-4 5
9 0

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.

only 4 cases

Edited by author 05.12.2014 19:38

Edited by author 05.12.2014 19:39

found my mistake. sorry for this post.

Edited by author 05.12.2014 22:16

Edited by author 05.12.2014 22:16