Codeforces Round #729 ABCD Solutions (Java/C++)
A. Odd Set Solution: Just need count the number of even numbers. If and only if the number of even numbers and the number of odd numbers are the same, then we output "Yes" Code: Java C++…
Codeforces Round #729 Priority Queue Solution (Java/C++)
Solution: First, for each + operation, we only need count the number of subsequences which include this operation. Let us assume the i-th operation is +, and the value of it is a[i].…
Codeforces Round #729 Strange Function Solution (Java/C++)
Solution: First, we can note that: when i is odd number, f(i)=2.Then we consider the case that f(i)=3. Based on the definition, if and only if i is even number and i cannot be divisible by 3, then f(i)=3.…
Codeforces Round #728 (Div. 2) ABC Solutions (Java/C++)
A. Pretty Permutations Solution: Obviously, only the two neighbors need to exchange positions with each other.For the case where n is an odd number, only the last three bits need to be exchanged with each other.…
Codeforces Round #727 ABCD Solutions (Java/C++)
A. Contest Start Solution: Each participant will make the following $\frac{t}{x}$ participants dissatisfied. But for last few participants, because no more participants following, so need subtract it.…
Codeforces Round #727 PriceFixed Solution (Java/C++)
Solution: For each product, we will buy exactly $a_i$ items. We will not buy more items. So, the total number items we buy will not change (because…
Codeforces Round #726 ABCD Solutions (Java/C++)
A. Arithmetic Array Solution: For the sum less than n, we just need append one number to make the sum equal to n. For the sum more than n, we append some zero to make n equal to the sum.…
Codeforces Round #726 Deleting Divisors Solution (Java/C++)
Solution: We can separate the number to 3 categories: odd, even but not include $2^n$ and $2^n$. Then we can get the state transition brought by one step operation:…
