牛客多校第六场 J Upgrading Technology dp

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牛客多校第六场 J Upgrading Technology dp

题意: 有n个技能,一开始都是0级,第i个技能从j-1级升到j级,花费$c_{i,j}$,但是花费不一定是正的所有的技能升到j级时,奖励$d_j$但是奖励也不一定是正的题解: 用sum[i][j]储存-c[i][j]的前缀和,即技能i升到j级后总共的收益. 再用w[j]储存f[j]的前缀和,代表所有的技能都升到j的收益. 再开一个数组maxxx[i][j]用于储存技能i至少升到j级的收益,即max(sum[i][j~m]). 然后枚举j,计算$w_j+\sum_{i=1}^{n}maxxx[…

2019 牛客多校第六场 J Upgrading Technology

题目链接:https://ac.nowcoder.com/acm/contest/886/J 题目大意略. 分析见代码. 代码如下 #include <bits/stdc++.h> using namespace std; #define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); #define Rep(i,n) for (int i = 0; i < (n); ++i) #define For(i,s,t)…

牛客多校第六场 J Heritage of skywalkert 随即互质概率 nth_element（求最大多少项模板）

链接:https://www.nowcoder.com/acm/contest/144/J来源:牛客网 skywalkert, the new legend of Beihang University ACM-ICPC Team, retired this year leaving a group of newbies again. Rumor has it that he left a heritage when he left, and only the one who has at lea…

牛客多校第3场 J 思维+树状数组+二分

牛客多校第3场 J 思维+树状数组+二分传送门:https://ac.nowcoder.com/acm/contest/883/J 题意: 给你q个询问,和一个队列容量f 询问有两种操作: 0.访问操作,如果询问的name在队列中存在的话,那么就输出队列中name对应的val值,然后将队列中name对应的元素扔到队列的尾部去,否则就直接将该元素插入到队列的尾部去 1.插入操作,得到队列中对应的name元素的v值为k,查询第k+v个元素的v值是多少题解: 已知,对于插入操作,我们需要很快的查询…

牛客多校第五场 J：Plan

链接:https://www.nowcoder.com/acm/contest/143/J 来源:牛客网时间限制:C/C++ 1秒,其他语言2秒空间限制:C/C++ 262144K,其他语言524288K 64bit IO Format: %lld 题目描述 There are n students going to travel. And hotel has two types room:double room and triple room. The price of a double…

牛客多校第六场-H-Pair

链接:https://ac.nowcoder.com/acm/contest/887/H来源:牛客网题目描述 Given three integers A, B, C. Count the number of pairs <x ,y> (with 1≤x≤Aand1≤y≤B) such that at least one of the following is true: - (x and y) > C - (x xor y) < C ("and", &q…

牛客多校第六场 C Generation I 组合数学阶乘逆元模板

链接:https://www.nowcoder.com/acm/contest/144/C来源:牛客网 Oak is given N empty and non-repeatable sets which are numbered from 1 to N. Now Oak is going to do N operations. In the i-th operation, he will insert an integer x between 1 and M to every set inde…