【codeforces 566G】Max and Min 计算几何 题意:平面上有一个在第一象限的点(x,y),现在两个人(max 和 min) 分别有n,m个正向量,两个人轮流操作,max的操作是选择一个向量,使得点加上这个向量(可以选择多次),min则是减。如果某一刻这个点x<0, y < 0,min赢,反之max赢,问你哪个人赢。 首先,假如说有...
a1[1],...a1[k-1] . . . an-1[0],an-1[1], ... an-1[k-1] Now a Set of size n is constructed by taking any value randomly from each of the arrays. e.g one such set can be {a0[0],a1[3],a2[2],... an-1[k-1]} My goal is to find out the min and max eleme...
2 +15 I think that this data structure is required for https://codeforces.com/problemset/problem/436/F which can be solved with sqrt decomposition and convex hull trick → Reply Jester 6 years ago, # ^ | +5 Thanks a lot! Found the problem very helpful. → Reply ...
Note In the first test case,S={0,1,3,4}S={0,1,3,4},a=mex(S)=2a=mex(S)=2,b=max(S)=4b=max(S)=4,⌈a+b2⌉=3⌈a+b2⌉=3. So33 is added intoSS, andSS becomes{0,1,3,3,4}{0,1,3,3,4}. The answer is44. In the second test case,S={0,1,4}S={0,...
Maxim is good at math and he suspects that the total result doesn't only depend on his maximum speedv, but also on his choice of the initial pointbi. Now Maxim is asking you to calculate for each ofncompetitions the minimum possible time that can be measured by the time counter. The ...
Two kittens, Max and Min, play with a pair of non-negative integers x and y. As you can guess from their names, kitten Max loves to maximize and kitten Min loves to minimize. As part of this game Min wants to make sure that both numbers, x and y became negative at the same time...
insert the minimum of xx and yy into the permutation in such a way that it becomes the first element; insert the maximum of xx and yy into the permutation in such a way that it becomes the last element. For example, if p=[1,5,4,2,3]p=[1,5,4,2,3] and we want to apply...
Min-Max 容斥+莫比乌斯反演,推式子 hot tea 洛谷题面传送门 hot tea. 首先注意到这个\(\text{lcm}\)特别棘手,并且这里的\(k\)大得离谱,我们也没办法直接枚举每个质因子的贡献来计算答案。不过考虑到如果我们把这里的\(\text{lcm}\)改为\(\gcd\)那么一遍莫比乌斯反演即可搞定,因此考虑将这里的\(\text{lcm...
其中max(dp[i-1][k])表示前 j - 1项 选取 i-1段以 k 结尾的最大值。 实现这个状态转移需要三层循环,限时 1 s,需要优化 我们发现最内层循环 for(int k=i-1;k<=j-1;k++) dp[i][j] = max(dp[i][j-1] + a[j], dp[i-1][k] + a[j] ) ...
LL minw[maxn][maxn], ans[maxn][maxn]; vector<PII> vertex; class MinMaxMax { public: long long findMin(vector<int> a, vector<int> b, vector<int> w, vector<int> v) { int n = SZ(v), m = SZ(a); REP(i, 0, n) REP(j, 0, n) minw[i][j] = ans[i][j] = INF...