## Codeforces Round #765 ABC Solutions (Java/C++)

A. Ancient Civilization Solution: Count the number of Xs where each bit is 1. If a bit has more than half of 1's, then y is 1 in that bit, otherwise it's 0.…

A. Ancient Civilization Solution: Count the number of Xs where each bit is 1. If a bit has more than half of 1's, then y is 1 in that bit, otherwise it's 0.…

A. Plus One on the Subset Solution: Obviously, output the maximum value minus the minimum value. Code: Java C++…

Solution: Trie + DP。 First, any number greater than 3 can be represented as the sum of 2s and 3s. So we have each of our segments of length either 2 or 3.…

A. Robot Cleaner Solution: Just simulate the behavior of the robot according to the requirements of the problem. Code: Java C++…

A. Square String? Solution: First, see whether the length of the string can be divisible by 2, and then verify whether the first half and the second half are the same.…

A. Forbidden Subsequence Solution: You can sort the strings directly, that is, you can directly output the string in the form of aaaabbbbccccdefg.…

A. Polycarp and Sums of Subsequences Solution: We assume a[1]<a[2]<a[3]. So naturally b[7]=a[1]+a[2]+a[3], b[6]=a[2]+a[3], b[5]=a[1]+a[3].…

Solution: First, we can directly consider all items together. Let's consider the following example (I modified it a little bit based on the sample data):…