Hello codeforces community , I am facing with that problem . For the input 12 how output is 5 ?
My solution is :
at time 0 we jump from 0 to 1 co-ordinate
at time 1 we jump from 1 to 3 co-ordinate
at time 2 we jump from 3 to 6 co-ordinate
at time 3 we dont jump because now jump length will be 4 , since we are on co-ordinate 6 so if we jump we will be on 6+4=10 co-ordinate ,
at time 4 we dont jump because now jump length will be 5 , since we are on co-ordinate 6 so if we jump we will be on 6+5=10 co-ordinate ,
at time 5 we will jump because now jump length will be 6 , since we are on co-ordinate 6 so if we jump we will be on 6+6=12 co-ordinate . We reached on co-ordinate 12 at time 6 . because one jump takes one second (5+1=6) .
So ans should be 6 . Why 5 ?
thanks :) . How to solve this problem if they said Find minimum jump and minimum time to reach X . Here we have to minimize the time and minimize the jump also . For example : 12 ans should be : minimum jump=4 and minimum time=5
Minimum number of jumps is always 1 :)
By the way, If you mean minimum number of jumps in minimum time (also for 12 the answer would be 3; 3+4+5=12), I think greedy is OK. Am I wrong?
Great solution . Thanks :) .
Your solution is correct i think :
16
time=6 , jump=4 (6+5+4+1=16)
24
time=7 , jump=5 (7+6+5+4+3=24)
17
minimum time=6 , minimum jump=4 (6+5+4+2)
How to implement it ? Is a linear loop is enough ? ( For minimum jump calculation)
int cnt=0,tot=0;
for(int i=time;i>=1;i--) {
tot+=i; cnt++; if(tot==x) break; if(tot>x) { tot-=i; cnt--; }}
cout<<cnt<<'\n';
17
minimum time=6 , minimum jump=4 (6+5+4+2)
am i wrong ?
I'm not totally sure about the algorithm, But your implementation is correct.
17
minimum time=6 , minimum jump=4 (6+5+4+2) is this correct too ?
Yes.