链接:https://codeforces.com/contest/1173/problem/C 题意: Nauuo is a girl who loves playing cards. One day she was playing cards but found that the cards were mixed with some empty ones. There are nn cards numbered from 11 to nn, and they were mixed with…
链接:https://codeforces.com/contest/1173/problem/B 题意: Nauuo is a girl who loves playing chess. One day she invented a game by herself which needs nn chess pieces to play on a m×mm×mchessboard. The rows and columns are numbered from 11 to mm. We denote…
链接:https://codeforces.com/contest/1173/problem/A 题意: Nauuo is a girl who loves writing comments. One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes. It's known that there were xx persons who would upvote…
D. Nauuo and Circle •参考资料 [1]:https://www.cnblogs.com/wyxdrqc/p/10990378.html •题意 给出你一个包含 n 个点的树,这 n 个点编号为 1~n: 给出一个圆,圆上放置 n 个位置,第 i 个位置对应树中的某个节点,并且不重复: 求在圆上还原这棵树后,使得边不相交的总方案数: •题解 ①为何每一颗子树一定是连续的一段圆弧? 假设不是连续的圆弧,如图所示: 为了使 x 接到树上,必然会有 x-y 或 x-z 相连的边,这样…
Codeforces Round #564 (Div. 1) A Nauuo and Cards 首先如果牌库中最后的牌是\(1,2,\cdots, k\),那么就模拟一下能不能每次打出第\(k+i\)张牌. 然后考虑每一张牌打出后还要打多少张牌以及这张牌是什么时候入手的,分别记为\(f_i,g_i\),那么答案就是\(f_i+g_i\)的最大值. #include<bits/stdc++.h> #define qmin(x,y) (x=min(x,y)) #define qmax(x,y)…
题目传送门 /* 题意:两堆牌,每次拿出上面的牌做比较,大的一方收走两张牌,直到一方没有牌 queue容器:模拟上述过程,当次数达到最大值时判断为-1 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string> #include <stack> #include <cmath> #inc…
传送门 参考资料 [1]: the Chinese Editoria A. Nauuo and Votes •题意 x个人投赞同票,y人投反对票,z人不确定: 这 z 个人由你来决定是投赞同票还是反对票: 判断 x 与 y 的相对大小是否确定? •题解 如果 x == y && z == 0,输出 '0': 如果 x-y > z,输出 '+': 如果 y-x > z,输出 '-': 反之,输出 '?': •Code #include<bits/stdc++.h> u…
B. Nauuo and Chess 题目链接:http://codeforces.com/contest/1173/problem/B 题目 Nauuo is a girl who loves playing chess. One day she invented a game by herself which needs n chess pieces to play on a m×m chessboard. The rows and columns are numbered from 1 t…
A. Nauuo and Votes 题目链接:http://codeforces.com/contest/1173/problem/A 题目 Nauuo is a girl who loves writing comments. One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes. It's known that there were xpersons…
E. Vladik and cards 题目链接 http://codeforces.com/contest/743/problem/E 题面 Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front of himself. Every card has a positive integer number not e…
C. Soldier and Cards Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/546/problem/C Description Two bored soldiers are playing card war. Their card deck consists of exactly n cards, numbered from 1 to n, all values are diff…
题目链接:http://codeforces.com/problemset/problem/546/C 题解: 用两个队列模拟过程就可以了. 特殊的地方是:1.如果等大,那么两张牌都丢弃 : 2.如果操作了很多次仍不能分出胜负,则认为平手.(至于多少次,我也不知道,只能写大一点碰运气,但要防止超时) 代码如下: #include<iostream>//C - Soldier and Cards #include<cstdio> #include<cstring> #in…
E. Vladik and cards time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vladik was bored on his way home and decided to play the following game. He took n cards and put them in a row in front…
E. George and Cards   George is a cat, so he loves playing very much. Vitaly put n cards in a row in front of George. Each card has one integer written on it. All cards had distinct numbers written on them. Let's number the cards from the left to the…
#include <iostream> using namespace std; int main(){ int n,x; cin >> n >> x; ; ; i < n ; ++ i){ int number; cin>> number; sum +=number; } ) sum = -sum; cout<<(sum%x ? sum/x+ : sum/x) <<endl; }…
题目链接: 题目 E. George and Cards time limit per test:2 seconds memory limit per test:256 megabytes 问题描述 George is a cat, so he loves playing very much. Vitaly put n cards in a row in front of George. Each card has one integer written on it. All cards had…
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 =…
Codeforces Round #383 (Div. 2) A. Arpa's hard exam and Mehrdad's naive cheat 题意 求1378^n mod 10 题解 直接快速幂 代码 #include<bits/stdc++.h> using namespace std; long long quickpow(long long m,long long n,long long k) { long long b = 1; while (n > 0) { if…