600A - Extract Numbers

This is a technical problem. You should do exactly what is written in problem statement.

600B - Queries about less or equal elements

Let's sort all numbers in a. Now let's iterate over elements of b and for element b_j find the index of lowest number that is greater than b_j. We can do that using binary search. That index will be the answer for value b_j.

Complexity: O(nlogn).

600C - Make Palindrome

Let's denote cnt_c — the number of occurences of symbol c. Let's consider odd values cnt_c. Palindrome can not contain more than one symbol c with odd cnt_c. Let's denote symbols with odd cnt_c as a₁, a₂...a_k (in alphabetical order). Let's replace any one of symbols a_k with symbol a₁, a_k - 1 with a₂ and so on until the middle of a. Now we have no more than one odd symbol. If we have some let's place it in the middle of the answer. First half of answer will contain occurences of symbol c in alphabetical order. The second half will contain the same symbols in reverse order. For example for string s = aabcd at first we will replace d by

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Compexity: O(n).

600D - Area of Two Circles' Intersection

If the circles don't intersect than the answer is 0. We can check that case with only integer calculations (simply by comparing the square of distance between centers with square of the sum of radiuses). If one of the circles is fully in other then the answer is the square of the smaller one. We can check this case also with only integer calculations (simply by comparing the square of distance between centers with square of the difference of radiuses).

So now let's consider the general case. The answer will be equal to the sum of two circular segments. Let's consider the triangle with apexes in centers if circles and in some intersecting point of the circles. In that triangle we know all three sides so we can compute the angle of the circular segment. So we can compute the square of circular sector. And the only thing that we should do now is to subtract the square of triangle with apexes in the center of circle and in the intersecting points of circles. We can do that by computing the half of absolute value of cross product. So we have the following formulas:

,

where d is the distance between centers of the circles. And also we should do the same thing with second circle by replacing of indices 1 ≤ ftrightarrow2.

Complexity: O(1).