A. Sum of 2050
Solution:
Obviously $n$ must be divisible by 2050. Then every decimal digit of $\frac n {2050}$ should be composed of the corresponding $10^k$. So just find the sum of each digit of $\frac n {2050}$.
Code:
B. Morning Jogging
Solution:
Obviously, Sort all the roads. Assign the first $m$ roads to the $m$ individuals in turn. The order of other roads doesn't matter.
Code:
Tips for implementation:
It is very common to sort all edges by length. But the hard thing is we need to put the roads back with some roads are already assigned to someone.
A personal way of implementation is: after assigned first $m$ roads. I sort the road by the index of checkpoints and index of the road again.
C. Fillomino 2
Solution:
Refer to the figure below, because the red part has been used up, the blue on the left of the red is the value of the upper one, and the green on the right of the red is the value of the right.
In the same way, the second time is when the number of times of 2 is exhausted, the same method can build the next layer.
Code:
D. Explorer Space
Just a normal DP problem.
E. Group Photo
Just a simple binary search problem. But there are lots of different cases that have to be considered. So more a implementation question.
