题目大意:给你一个时间(hh:mm),求最少经过多少分钟才能使这个时间变成回文. 解题思路:模拟,先判断0的情况,然后每过1分钟判断一次即可. C++ Code: #include<cstdio> int main(){ int h,m; scanf("%d:%d",&h,&m); if(h==m%10*10+m/10){ puts("0"); return 0; } for(int i=1;;++i){ ++m; if(m==60){…

CodeForces 816B Karen and Coffee(前缀和,大量查询) Description Karen, a coffee aficionado, wants to know the optimal temperature for brewing the perfect cup of coffee. Indeed, she has spent some time reading several recipe books, including the universally ac…

[题目链接]:http://codeforces.com/contest/816/problem/A [题意] 让你一分钟一分钟地累加时间; 问多长时间以后是个回文串; [题解] reverse之后如果和原串相同,则为回文串; 模拟就好 [Number Of WA] 0 [完整代码] #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|…

A. Karen and Morning 传送门:http://codeforces.com/contest/816/problem/A 水题,参考程序如下: #include <stdio.h> int main(void) { int h, m, H, M; scanf("%d:%d", &h, &m); ; i <= * ; i++) { H = (h + (m + i) / ) % ; M = (m + i) % ; == M % &&…

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…

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…

题目链接:http://codeforces.com/contest/816/problem/B 题目意思:给出 n 个recipes,第 i 个(1<= i <=n)recipes 表明 coffee 调制的推荐温度范围是 [li, ri] 之间.现在有 q 个问题,每个问题需要回答 coffee 在范围 [a, b] 之间,共有多少个数满足至少有 k 个推荐. 题目解析:这题主要是考我们对于大范围(最大200000),如何处理数据.方法是很容易想到的,但要考虑优化,即离线处理.20w *…

Problem Description On the way to school, Karen became fixated on the puzzle game on her phone! The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0. One move consists of…

题目链接 816B Karen and Coffee 题目分析 题意:有个人在学泡咖啡,因此看了很多关于泡咖啡温度的书,得到了n种推荐的泡咖啡温度范围[L1,R1] ,此人将有k种做法推荐的温度记为可用温度(个人翻译),然后给出q次询问,问区间[L2,R2]内的温度,有多少个温度是可用温度(每个整数代表一个温度) 思路:一开始用的是线段树写的,不过姿势不对,TLE了,然后改过来后,发现时间比较长,就考虑一下优化的方法. 比线段树某些功能更优的算法:差分思想,在对某一区间每个位置上的数加上一个值x…

LINK 思路 首先发现依赖关系是一个树形的结构 然后因为直接算花多少钱来统计贡献不是很好 因为数组开不下 那就可以算一个子树里面选多少个的最小代价就可以了 注意统计贡献的时候用优惠券的答案只能在1号点进行统计 //Author: dream_maker #include<bits/stdc++.h> using namespace std; //---------------------------------------------- //typename typedef long lon…