http://codeforces.com/contest/816/problem/B 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 r…

C. Karen and Game time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output On the way to school, Karen became fixated on the puzzle game on her phone! The game is played as follows. In each level,…

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…

http://codeforces.com/contest/816/problem/E 题意: 去超市买东西,共有m块钱,每件商品有优惠卷可用,前提是xi商品的优惠券被用.问最多能买多少件商品? 思路: 第一件商品使用优惠券不需要前提,别的都是需要的,然后这样就形成了一棵以1为根的树. 这样,很容易想到是树形dp. d[u][j][0/1]表示以u为根的子数中选择j件商品所需的最少花费,0/1表示u商品是否能用优惠券. 解释一下代码中的sz[],它所代表的是以u为根的子树的结点数. 当我们现在访…

http://codeforces.com/contest/816/problem/A 题意: 给出一个时间,问最少过多少时间后是回文串. 思路: 模拟,先把小时的逆串计算出来: ① 如果逆串=分钟,那么此时已经是回文串了. ② 如果逆串>分钟,那么只需要逆串-分钟即可.(注意此时逆串>=60的情况) ③ 如果逆串<分钟,此时在这个小时内要构成回文串已经是不可能的了,那么就加上60-minute分钟,进入一个新的小时,然后重复上述步骤. #include<iostream>…

C. Karen and Supermarket     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.…

打个表出来看看,其实很明显. 推荐打这俩组 11 1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000 10000000000 12 1 10 100 1000 10000 100000 1000000 10000000 100000000 1000000000 10000000000 100000000000 打出表来看出来,n为偶数时,每隔两行,对原序列的奇数项分配的权重形成二项展开式. n为奇数时,每隔四行,形成二…

容易发现,删除的顺序不影响答案. 所以可以随便删. 如果行数大于列数,就先删列:否则先删行. #include<cstdio> #include<algorithm> using namespace std; int p1,ans1[510*110],ans2[510*110],p2; int n,m,a[110][110]; int main(){ scanf("%d%d",&n,&m); for(int i=1;i<=n;++i){ f…

1.题目A:Karen and Morning 题意: 给出hh:mm格式的时间,问至少经过多少分钟后,该时刻为回文字符串? 思路: 简单模拟,从当前时刻开始,如果hh的回文rh等于mm则停止累计.否则,根据rh和mm的大小来累计sum,然后hh+1,不断尝试. #include<iostream> using namespace std; int main() { int hh,mm; char c; while (cin >> hh >> c >> mm…

传送门:Problem B https://www.cnblogs.com/violet-acmer/p/9721160.html 题意: Karen有n个关于煮咖啡的食谱,每个食谱都有个煮咖啡的最适宜的温度范围,给你 m 个操作,每个操作都是个温度区间,问在此区间内满足条件的温度个数. 需要满足的条件为:在温度 T 时,在n个食谱中查找最宜温度包含T的食谱个数,只有当食谱个数 >= k 时,此温度才算可以计入答案. 题解: 差分数组 AC代码: #include<iostream> #…

A. Karen and Morning time limit per test 2 seconds memory limit per test 512 megabytes input standard input output standard output Karen is getting ready for a new school day! It is currently hh:mm, given in a 24-hour format. As you know, Karen loves…

上紫啦! E题1:59压哨提交成功翻盘 (1:00就做完了调了一个小时,还好意思说出来? (逃)) 题面太长就不复制了,但是配图很可爱所以要贴过来 九条可怜酱好可爱呀 A - Karen and Morning 询问从当前时刻过多久,时间会形成回文串的形式. 暴力呀暴力 #include<iostream> #include<algorithm> #include<cstring> #include<cstdio> #include<cmath>…

A-C传送门 D Karen and Cards 技巧性很强的一道二分优化题 题意很简单 给定n个三元组,和三个维度的上限,问存在多少三元组,使得对于给定的n个三元组中的每一个,必有两个维度严格小于. 首先我们根据一个维度(c维)对n个三元组排序,然后枚举答案在这个维度的取值. 此时序列被分成了两个部分,前半部分 满足所有c大于等于i 后半部分满足所有c严格小于i(即已有一个维度小于) 通过累计,我们知道此时前半部a维的最大值ma和b维的最大值mb. 显然可能存在的三元组答案,必然首先满足a维和…

1. 815A Karen and Game 大意: 给定$nm$矩阵, 每次选择一行或一列全部减$1$, 求最少次数使得矩阵全$0$ 贪心, $n>m$时每次取一列, 否则取一行 #include <iostream> #include <sstream> #include <algorithm> #include <cstdio> #include <cmath> #include <set> #include <ma…

A:暴力枚举第一列加多少次,显然这样能确定一种方案. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 110 char getc(){char c=getchar()…

python 2.7,用来熟悉Python 由于都是智障题,所以我也不讲述题意和题解,直接贴代码了-- A import sys h,m = map(int,raw_input().split(":")) ans = 0 while True: if h%10 == m/10 and h/10 == m%10: break ans = ans + 1 h,m = (h+m/59)%24,(m+1)%60 print ans B import sys maxn = 200005 # cl…

Codeforces Round #439 (Div. 2) codeforces 869 A. The Artful Expedient 看不透( #include<cstdio> int main(){ puts("Karen"); ; } 15ms codeforces 869B. The Eternal Immortality(数学,水) 题意:输出两个数的阶乘的商的 个位数 题解:两数之差大于5,个位数就是0.小于5直接个位相乘即可. #include<cs…

Codeforces Round #366 (Div. 2) A I hate that I love that I hate it水题 #I hate that I love that I hate it n = int(raw_input()) s = "" a = ["I hate that ","I love that ", "I hate it","I love it"] for i in ran…

Codeforces Round #354 (Div. 2) Problems     # Name     A Nicholas and Permutation standard input/output 1 s, 256 MB    x3384 B Pyramid of Glasses standard input/output 1 s, 256 MB    x1462 C Vasya and String standard input/output 1 s, 256 MB    x1393…

直达–>Codeforces Round #368 (Div. 2) A Brain’s Photos 给你一个NxM的矩阵,一个字母代表一种颜色,如果有”C”,”M”,”Y”三种中任意一种就输出”#Color”,如果只有”G”,”B”,”W”就输出”#Black&White”. #include <cstdio> #include <cstring> using namespace std; const int maxn = 200; const int INF =…

 cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅.....       其实这个应该是昨天就写完的,不过没时间了,就留到了今天.. 地址:http://codeforces.com/contest/651/problem/A A. Joysticks time limit per test 1 second memory limit per test 256…

Codeforces Round #279 (Div. 2) 做得我都变绿了! Problems     # Name     A Team Olympiad standard input/output 1 s, 256 MB  x2377 B Queue standard input/output 2 s, 256 MB  x1250 C Hacking Cypher standard input/output 1 s, 256 MB  x740 D Chocolate standard in…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

CF469 Codeforces Round #268 (Div. 2) http://codeforces.com/contest/469 开学了,时间少,水题就不写题解了,不水的题也不写这么详细了. A 水题 //#pragma comment(linker, "/STACK:102400000,102400000") #include<cstdio> #include<cmath> #include<iostream> #include<…

题目传送门 /* 贪心 + 模拟:首先,如果蜡烛的燃烧时间小于最少需要点燃的蜡烛数一定是-1(蜡烛是1秒点一支), num[g[i]]记录每个鬼访问时已点燃的蜡烛数,若不够,tmp为还需要的蜡烛数, 然后接下来的t秒需要的蜡烛都燃烧着,超过t秒,每减少一秒灭一支蜡烛,好!!! 详细解释:http://blog.csdn.net/kalilili/article/details/43412385 */ #include <cstdio> #include <algorithm> #i…

题目传送门 /* 题意:从前面找一个数字和末尾数字调换使得变成偶数且为最大 贪心:考虑两种情况:1. 有偶数且比末尾数字大(flag标记):2. 有偶数但都比末尾数字小(x位置标记) 仿照别人写的,再看自己的代码发现有清晰的思维是多重要 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include…

#include <iostream> #include <string> using namespace std; int main(){ int n; cin >> n; string str; cin >> str; , x = ; ; i < n ; ++ i){ if(str[i] == 'B') cnt+=(x << i); } cout<<cnt<<endl; }   Codeforces Round…

Codeforces Round #160 (Div. 1) A - Maxim and Discounts 题意 给你n个折扣,m个物品,每个折扣都可以使用无限次,每次你使用第i个折扣的时候,你必须买q[i]个东西,然后他会送你{0,1,2}个物品,但是送的物品必须比你买的最便宜的物品还便宜,问你最少花多少钱,买完m个物品 题解 显然我选择q[i]最小的去买就好了 代码 #include<bits/stdc++.h> using namespace std; const int maxn =…