Educational Codeforces Round #105 Dogeforces - Disjoint set + Construction
Solution: Sort the salary. We build a dynamic disjoint set. In the beginning, there only be n lower-level employees. And the father of these employees is empty.…
Educational Codeforces Round #105 1D Sokoban - Binary Search
Solution: Classic binary search problem. We only consider the positive part. The negative part can be similar.…
Codeforces Round #718 Explorer Space - DP
Solution: First of all, if $k$ is not an even number, there must be no answer. Define $map[i][j][d]$ (where $0 \leq d <4$) represents the side length of $(i,j)$ in four directions.…
Codeforces Round #718 Group Photo - Binary Search
Solution: First of all, Base on ci−ci−1≤ci+1−ci and pi−pi−1≥pi+1−pi we can roughly found that：…
Codeforces Round #718 ABCDE Solutions
A. Sum of 2050 Solution: Obviously $n$ must be divisible by 2050. Then every decimal digit of $\frac n {2050}$ should be composed of the corresponding...…
Codeforces Round #717 Cut - Number theory + Binary search + DP
Solution: First of all, "The product of every subarray is equal to its LCM" means "All these numbers are mutually prime". "All these numbers are mutually prime" means "In all these numbers, the prime factors of any two numbers are different".…
Codeforces Round #717 Baby Ehab Partitions Again - Math+01 Knapsack problem
Solution Define $sum=\sum_{i=1}^n a_i$. If $sum$ is a odd number, then of course no need remove anything.…
Codeforces Round #717 AGAGA XOOORRR - bitmasks
Solution: If the XOR sum of all numbers is 0. Then it must can make it. Because if and only if a[1]⊕a[2]⊕...⊕a[n−1]=a[n], the XOR sum is 0. So we can make the array as: (a[n],a[n]).…
