To stay woke and attentive during classes, Karen needs some coffee! 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 u…

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…

To stay woke and attentive during classes, Karen needs some coffee! 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 u…

The Cartesian coordinate system is set in the sky. There you can see n stars, the i-th has coordinates (xi, yi), a maximum brightness c, equal for all stars, and an initial brightness si (0 ≤ si ≤ c). Over time the stars twinkle. At moment 0 the i-th…

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 *…

传送门:http://codeforces.com/problemset/problem/612/D (转载请注明出处谢谢) 题意: 给出数字n和k,n表示接下来将输入n个在x轴上的闭区间[li,ri],找出被包含了至少k次的点,并求由这些点组成的连续区间的数目,并使该数目最小.输出该数目并将区间从左到右(x的正方向)输出. 比如样例1,给出区间[0,5],[-3,2],[3,8],那么被覆盖了至少两次的区间就是[0,2],[3,5],有两个. 解题思路: 处理区间覆盖,一上来就想到了前缀和,普…

[题目链接]:http://codeforces.com/contest/816/problem/B [题意] 给你很多个区间[l,r]; 1<=l<=r<=2e5 一个数字如果被k个以上的区间覆盖到; 则称之为特殊数字; 给你q个询问区间; 问你这q个区间里面 每个区间里面有多少个特殊数字; [题解] 对于覆盖的区间l,r add[l]++,sub[r+1]++, 然后顺序枚举 now+=add[i]; now-=sub[i]; 如果now>=k则i是一个特殊数字; 写个前缀和O…

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…

大意: 给定n*m初始字符矩阵, 有k个新矩阵, 第$i$个矩阵是由初始矩阵区间赋值得到的, 求选择一个新矩阵, 使得其余新矩阵到它距离和最小. 字符集比较小, 可以考虑每次区间覆盖对每个字符的贡献. 区间覆盖转化为差分, 然后前缀和优化. 刚开始辅助数组开多了, 卡内存卡了好久. 看别人代码似乎可以再优化掉两个辅助数组. #include <iostream> #include <sstream> #include <algorithm> #include <c…

A /* Huyyt */ #include <bits/stdc++.h> using namespace std; typedef long long ll; ]; ]; map<ll, int> mp; int main() { int n; cin >> n; ; i <= n; i++) { cin >> a[i]; } ; ; i <= n; i++) { memset(num, , sizeof(num)); ; ; j <…