Hi friends it's not a professional Editorial I just wrote it myself and i hope it can help you.↵
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A:↵
you see the only way we can decrease number of prime factors 2 is to divide n by 2, and the same for 3 and 5;↵
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so for decreasing prime a factor 5 we have to use operation 3 and after that two times operation 1.↵
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for decreasing prime a factor 3 we have to use operation 2 and after that operation 1.↵
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so if we define f(x) to be number of prime factors x in n answer will be f(2) + 2 * f(3) + 3 * f(5);↵
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B:↵
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for each query let cnt[x] (0 <= x && x < 3) be number of elements like y that y mod 3 = x.↵
then easily we can prove that we chould not do anything with numbers divisible by 3. and for cnt[1] and cnt[2] we can merge min(cnt[1], cnt[2]) this operation "merge a number x(x mod 3 = 1) with a number y(y mod 3 = 2)" that these make numbers divisible by 3 again and for remaining numbers that all are equal mod 3 if they are x elements we can make [x / 3] numbers divisible by 3 with them↵
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C:↵
imagine numbers arr [0, 1, 2, 3, 4, 5] now each round find first 0 1 2 3 4 5 as a subsequence greedy this way "find first 0 after that first 1 and so on" then put them in a group and don't use them in next rounds.↵
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whenever you couldn't find any new subsequence you must delete all remaining elements↵
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D:↵
first of all let's call prime numbers in a type_1, others in a type_2, and primes in b type_3, and others in b type_4↵
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now let's solve problem recursive see biggest number(x).↵
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if it is prime it's type_3 cause for each i (i >= 2) Pi > i so if it's type_1 we should have a bigger prime in our numbers that it is not true. so now delete x and y(that P(y) = x) and solve remaining elements recursive.↵
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else it's type_2 cause if it's type_4 imagine it's type_2 so we have inserted x / y that (y > 1) in b that it's false cause x is biggest number. so now delete x and y(maximum divisor of x that is smaller than x) and solve remaining elements recursive.↵
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E:↵
let's make a dfs and solve problem with dfs tree.↵
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these tree is bipartite and as we have n >= 2 for each vertex v, d(v) > 1 so if we choose any of these parts we will have condition that we need.↵
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now let's choose smaller one.↵
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F:↵
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I suggest you read [my_solution](https://codeforces.net/contest/1176/submission/55360118) for problem F(Destroy it!). I have dp[N][10] that dp[i][j] is maximum sum if we have get x cards from first i round and x mod 10 = j. and it's easy to see that in each round we may use only 3 strongest card that cost 1, strongest card cost 2 and strongest card cost 3 so by a mask on these(at most 5)cards you can update dp easily.↵
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I HOPE IT WAS USEFUL FOR YOU :)
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A:↵
you see the only way we can decrease number of prime factors 2 is to divide n by 2, and the same for 3 and 5;↵
↵
so for decreasing prime a factor 5 we have to use operation 3 and after that two times operation 1.↵
↵
for decreasing prime a factor 3 we have to use operation 2 and after that operation 1.↵
↵
so if we define f(x) to be number of prime factors x in n answer will be f(2) + 2 * f(3) + 3 * f(5);↵
↵
B:↵
↵
for each query let cnt[x] (0 <= x && x < 3) be number of elements like y that y mod 3 = x.↵
then easily we can prove that we chould not do anything with numbers divisible by 3. and for cnt[1] and cnt[2] we can merge min(cnt[1], cnt[2]) this operation "merge a number x(x mod 3 = 1) with a number y(y mod 3 = 2)" that these make numbers divisible by 3 again and for remaining numbers that all are equal mod 3 if they are x elements we can make [x / 3] numbers divisible by 3 with them↵
↵
C:↵
imagine numbers arr [0, 1, 2, 3, 4, 5] now each round find first 0 1 2 3 4 5 as a subsequence greedy this way "find first 0 after that first 1 and so on" then put them in a group and don't use them in next rounds.↵
↵
whenever you couldn't find any new subsequence you must delete all remaining elements↵
↵
D:↵
first of all let's call prime numbers in a type_1, others in a type_2, and primes in b type_3, and others in b type_4↵
↵
now let's solve problem recursive see biggest number(x).↵
↵
if it is prime it's type_3 cause for each i (i >= 2) Pi > i so if it's type_1 we should have a bigger prime in our numbers that it is not true. so now delete x and y(that P(y) = x) and solve remaining elements recursive.↵
↵
else it's type_2 cause if it's type_4 imagine it's type_2 so we have inserted x / y that (y > 1) in b that it's false cause x is biggest number. so now delete x and y(maximum divisor of x that is smaller than x) and solve remaining elements recursive.↵
↵
E:↵
let's make a dfs and solve problem with dfs tree.↵
↵
these tree is bipartite and as we have n >= 2 for each vertex v, d(v) > 1 so if we choose any of these parts we will have condition that we need.↵
↵
now let's choose smaller one.↵
↵
F:↵
↵
I suggest you read [my_solution](https://codeforces.net/contest/1176/submission/55360118) for problem F(Destroy it!). I have dp[N][10] that dp[i][j] is maximum sum if we have get x cards from first i round and x mod 10 = j. and it's easy to see that in each round we may use only 3 strongest card that cost 1, strongest card cost 2 and strongest card cost 3 so by a mask on these(at most 5)cards you can update dp easily.↵
↵
I HOPE IT WAS USEFUL FOR YOU :)
