Let's iterate over the first number of the pair, let it be x. Then we need to count numbers from 1 to m with the remainder of dividing 5 equal to (5 - xmod5)mod 5. For example, you can precalc how many numbers from 1 to m with every remainder between 0 and 4.
Let's sort the array. Let cur = 1. Then walk through the array. Let's look at current number. If it is greater or equal to cur, then let's increase cur by 1. Answer is cur.
Let's do dfs. Suppose that we now stand at the vertex u. Let v be some ancestor of vertex u. Then dist(v, u) = dist(1, u) - dist(1, v). If dist(v, u) > au, then the vertex u makes v sad. So you must remove the whole subtree of vertex u. Accordingly, it is possible to maintain a minimum among dist(1, v) in dfs, where v is ancestor of u (vertex, in which we now stand). And if the difference between the dist(1, u) and that minimum is greater than au, then remove au with the whole subtree.
Находясь в состоянии d[i][j][cnt][end], можно добавить следующую букву в строках s или t, при том она не будет входить в ответную подпоследовательность. Тогда d[i + 1][j][cnt][0] = max(d[i + 1][j][cnt][0], d[i][j][cnt][end]), d[i][j + 1][cnt][0] = max(d[i][j + 1][cnt][0], d[i][j][cnt][end]). То есть новое значение end равно 0, поскольку новая буква не входит в ответную подпоследовательность.
Если s[i] = t[j], то, если end = 1, то можно обновить d[i + 1][j + 1][k][1] = max(d[i][j][k][end] + 1, d[i + 1][j + 1][k][1]). Поскольку end = 1, при добавлении элемента к ответной подпоследовательности, количество подстрок, из которых она состоит, останется таким же. Если end = 0, то можно обновить d[i + 1][j + 1][k + 1][1] = max(d[i][j][k][end] + 1, d[i + 1][j + 1][k + 1][1]). В этом случае новые символы, которые мы пытаемся добавить к ответной подпоследовательности, будут образовывать уже новую подстроку, поэтому в этом случае переход из состояния с k в состояние с k + 1.
Ответом будет являться наибольшее число среди состояний вида d[n][m][k][end], где значения k и end принимают всевозможные значения.
Let's use the method of dynamic programming. Let d[i][j][cnt][end] be answer to the problem for the prefix of string s of length i and for the prefix of string t of length j, we have chosen cnt substrings. end = 1, if both last characters of the prefixes are included in the maximum subsequence and end = 0 otherwise.
When the state is d[i][j][cnt][end], you can add the following letters in the string s or t, though it will not be included in the response subsequence. Then d[i + 1][j][cnt][0] = max(d[i + 1][j][cnt][0], d[i][j][cnt][end]), d[i][j + 1][cnt][0] = max(d[i][j + 1][cnt][0], d[i][j][cnt][end]). So the new value of end is 0, because the new letter is not included in the response subsequence.
If s [i] = t [j], then if the end = 1, we can update the d [i + 1] [j + 1] [k] [1] = max (d [i] [j] [k ] [end] + 1, d [i + 1] [j + 1] [k] [1]). Since the end = 1, adding an element to the response subsequence, the number of substring, of which it is composed, will remain the same. If end = 0, we can update the d [i + 1] [j + 1] [k + 1] [1] = max (d [i] [j] [k] [end] + 1, d [i + 1] [j + 1] [k + 1] [1]). In this case, the new code, which we try to add the subsequence response will have to form a new substring, so in this case the transition from a state k to the state k + 1 a.
The answer will be the largest number among the states of the d [n] [m] [k] [end], where the values of k and end take all possible values.
Let's find the triangle with maximum area among all triangles whose vertices belong to the set of points. (For N2 with the convex hull and the two pointers). We can prove that if we take the triangle, whose vertices are the midpoints of the sides of the triangle with maximum area, the area of such a triangle is not greater than 4S, and it contains all the points of the set. Let us assume that there is a point lying outside the triangle-response. Then we can get longer height to some side of triangle, so we have chosen a triangle with not maximum area(contradiction).
