Codeforces Round #748 Red-Black Number Solution (Java/C++)
Solution: A very obvious DP problem. It is not difficult to think of several states: 1. The first i numbers; 2. |r-l|; 3. The remainder of division by A; 4. The remainder of division by B.…
Educational Codeforces Round 115 ABCDE Solutions (Java/C++)
A. Computer Game Solution: Scan from left to right one by one. As long as one of the columns on the left can be reached, all safe cells in the current column can be reached.…
Codeforces Round #747 ABCDEF Solutions (Java/C++)
A. Consecutive Sum Riddle Solution: Obviously $\frac {(l+r)\cdot (r-l+1)} 2 = n$, so $(l+r)\cdot (r-l+1) = 2\cdot n$.…
Codeforces Round #746 ABCD Solutions (Java/C++)
A. Gamer Hemose Solution: Obviously, we only need to choose the two weapons with the highest attack power to attack alternately. Code: Java C++…
Codeforces Round #746 Bakry and Partitioning Solution (Java/C++)
Solution: First, if the XOR of each component is the same (represent by X), then the XOR of all nodes is either X or 0.…
Codeforces Round #746 Hemose in ICPC ? Solution (Java/C++)
Solution: First, because gcd(a,b,c) must be less than or equal to gcd(a,b) or gcd(b,c). Therefore, the maximum distance must be the largest side.…
Codeforces Round #745 (Div. 2) ABCDE Solutions (Java/C++)
A. CQXYM Count Permutations Solution: It is not difficult to think that there are x of i satisfying the condition in an permutations , then if we reverse the permutations, there are 2n-x of i satisfying the condition.…
Codeforces Round #745 (Div. 2) Train Maintenance Solution (Java/C++)
Solution: First of all, we can almost immediately notice that if x[i]>m, then this must have no effect, and we can ignore it directly. Therefore, the value range of x[i] is reduced from 0~$10^9$ to 0~$2\cdot 10^5$.…
