For every symbol we should determine how to rotate the disk. This can be done either by formula: min(abs(a[i] - b[i]), 10 - abs(a[i] - b[i])) or even by the two
for cycles: in both directions.
First count the number of marks that are less than y. If there are more than such marks, we can't satisfy the second condition (about the median), and the answer is -1. Otherwise we can get exactly such number of y marks so that the total number of marks greater than or equal to y is at least (maybe it's already satisfied). This is the required action for satisfying the second condition.
Now, in order not to break the first condition, get the remaining marks as lower as possible — all ones — and check the sum of the marks. If it is greater than x, the answer is -1, otherwise the correct answer is found.
There are three cases here, though some of them can be merged.
- If the start and finish cells are equal, let's count the intact neighbours of this cell. If there is one, move there and instantly move back — the answer is YES. Otherwise it's NO.
- If the start and finish cells are neighbours, the solution depends on the type of the destination cell. If it's cracked, the answer is YES — we can just move there and fall down. Otherwise it must have at least one intact neighbour to get the positive answer — we can move to the finish cell, then to this intact neighbour, and then return to the finish cell.
- In the general case, check if the path from the start cell to the finish cell exists. If it doesn't, the answer is NO. Otherwise check the type of the destination cell. If it's cracked, it must have at least one intact neighbour, and if it's intact, it must have two intact neighbours.
Let's count the values dp[r][s][p] — the probability of the situation when r rocks, s scissors and p papers are alive. The initial probability is 1, and in order to calculate the others we should perform the transitions.
Imagine we have r rocks, s scissors and p papers. Let's find the probability of the rock killing scissors (the other probabilities are calculated in the same way). The total number of the possible pairs where one species kills the other one is rs + rp + sp, and the number of possible pairs (rock, scissors) is rs. As all meetings are equiprobable, the probability we want to find is . This is the probability with which we go the the state dp[r][s — 1][p], with the number of scissors less by one.
In the end, for example, to get the probability of the event that the rocks are alive, we should sum all values dp[i] for i from 1 to r (the same goes to the other species).
At first find the position of each element which is used in swap (using map). Now let's find the answer. It consists of the two parts. First part is the number of inversions formed by only whose elements which took part in the swaps. They can be counted by one of the standard ways: mergesort or Fenwick tree. The second part is the number of inversions formed by pairs of elements where one element has been swapped even once, and the other element stayed at his position. Let's consider the following test:
2 2 6 4 8
The global sequence will look as follows: [1 6 3 8 5 2 7 4 9 ...], and here is the array of swapped elements: [6 8 2 4].
Let's understand with which numbers the number 8 forms the inversions. The only elements that could do that are the elements between the initial position of the number 8 (where the number 4 is now) and its current position: [5 2 7]. There are two numbers on this segment which didn't take part in swaps: 5 and 7. The number 2 should not be counted as it took part in the swaps and we have already counted it in the first part of the solution.
So we should take the count of numbers between 8's indices in the global sequence (8 - 4 - 1 = 3) and subtract the count of numbers between its indices in the swaps array (4 - 2 - 1 = 1). We'll get the number of inversions formed by the element 8 and the elements which haven't moved at all, it's 2. Counting this value for all elements which have been swapped at least once, we get the second part of the answer. All operations in the second part of the solution can be performed using sorts and binary searches.