//Codeforces - CF803C #include <bits/stdc++.h> typedef long long valueType; int main() { valueType N, K; std::cin >> N >> K; if (N / K < (K - 10) / 2) { std::cout << -1 << std::endl; return 0; } if (N < K * (K +
ei). Two intervals i1 and i2 are called nested if i2 lies completely inside of i1. Example:- (2,6) and (3,4) are nested, since (3,4) is a part of (2,6). Similarly k intervals i1, i2, i3...ik are called nested if, i2 lies inside i1, i3 lies inside ...
during solving the problem. Solution is quite big, and it may not be clear, because there is only a code. If you test it, you can see it's correct. I will share
SolutionSolution 令gcdgcd为xx,那么我们将整个序列/x/x,则序列的和就变成了nxnx,所以xx必定为nn的约数所以现在就是要构造出一个序列长度为kk,和为nxnx。我们令前k−1k−1个为1,2...k−11,2...k−1最后一个再用nx−nx−这k−1k−1个的和就是最后一个数了。最后的答案就是构造出来的序...
I have an N^4 DP solution. as follows. I choose a point i and sort all the other points around this point according to angle (similar to convex hull graham scan). Then, I write the following bottom-up dp formulation: DP[B][A] stores the best convex polygon with the last two point...