Codeforces 919 A. Supermarket】的更多相关文章

这场cf有点意思,hack场,C题等于1的特判hack很多人(我hack成功3个人,上分了,哈哈哈,咳咳...) D题好像是树形dp,E题好像是中国剩余定理,F题好像还是dp,具体的不清楚,最近dp的题目好多,一会滚去学dp. 写A,B,C的题解. A. Supermarket   time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard outpu…
题目描述 Given an integer xx . Your task is to find out how many positive integers nn ( 1<=n<=x1<=n<=x ) satisfy  where a,b,pa,b,p are all known constants. 输入输出格式 输入格式: The only line contains four integers a,b,p,xa,b,p,x ( 2<=p<=10^{6}+32<…
题目描述 You are given a graph with nn nodes and mm directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then th…
C. Seat Arrangements   time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Suppose that you are in a campus and have to go for classes day by day. As you may see, when you hurry to a classroom,…
  B. Perfect Number   time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output We consider a positive integer perfect, if and only if the sum of its digits is exactly 10. Given a positive integer k…
A B C #include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!!\n") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,…
On the way home, Karen decided to stop by the supermarket to buy some groceries. She needs to buy a lot of goods, but since she is a student her budget is still quite limited. In fact, she can only spend up to b dollars. The supermarket sells n goods…
On the way home, Karen decided to stop by the supermarket to buy some groceries. She needs to buy a lot of goods, but since she is a student her budget is still quite limited. In fact, she can only spend up to b dollars. The supermarket sells n goods…
Codeforces 815 C 考虑树型dp. \(dp[i][0/1][k]\)表示现在在第i个节点, 父亲节点有没有选用优惠, 这个子树中买k个节点所需要花的最小代价. 然后转移的时候枚举i的一个儿子u, 然后还要枚举在u的子树中选择了多少个节点l, 则\(dp[i][0/1][k+l]=dp[i][0/1][k]+dp[u][0/1][l]\). 还要注意转移顺序. 最后枚举最后一个\(dp[1][1][i]\leq limit\)的i就是答案.…
Karen and Supermarket 感觉就是很普通的树形dp. dp[ i ][ 0 ][ u ]表示在 i 这棵子树中选择 u 个且 i 不用优惠券的最小花费. dp[ i ][ 1 ][ u ]表示在 i 这棵子树中选择 u 个且 i 用优惠券的最小花费. 注意这个转移总的合起来是O(n ^ 2)的. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk…