Of course a segment tree problem. Let's start from $f$. Then actually, the state of last day is fixed. Also fixed for the day before last day. The only thing we need to know is how many zeros between $l$ to $r$. Then we can know what is changed.…
First let's choose any two numbers, $a_i$ and $a_j$. So There must be $-a_i$ and $-a_j$ in $a$. And so there is no need to consider the symbol. Let's assume $a_i>0,\ a_j>0$.…
Let current query is x, let remian=x mod d. If x>=10d}+remain. Then there must be a solution for x.…
Solutions for all problems in Codeforces Round #702 Div3.…
Solutions for C2, D and E2 in Codeforces Round #708 Div2.…
There is only one case to make product is a perfect square: For each number, the number of occurrences of prime factors is count[i]. if any different $i$, $j$ to make count[i]=count[j].…
This problem remind you about the memory limitation at very begin. Normally, it's State compression DP.…
When $k=3$, if n is a odd number, then the answer is: $a_0=1$, $a_1=\frac{n-1}{2}$, $a_2=\frac{n-1}{2}$ When $k>3$, we just fill up the array with 1 until last 3 elements.…
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