C. Famil Door and Brackets 题目连接: http://www.codeforces.com/contest/629/problem/C Description As Famil Door's birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of roun…

B. Far Relative's Problem 题目连接: http://www.codeforces.com/contest/629/problem/B Description Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them can come to the party in a specific range of…

题目链接: http://www.codeforces.com/contest/629/problem/E 题解: 树形dp. siz[x]为x这颗子树的节点个数(包括x自己) dep[x]表示x这个节点的深度,从1开始(其实从什么开始都可以,我们这里用到的只是相对距离) 对于查询u,v,总共有三种情况: 1.u为公共祖先 设x为(u,v)链上u的儿子,则我们知道新边只能从非x子树的点(n-siz[x]连到以v为根的子树上的点(siz[v]) 则新边的总条数为(n-siz[x])*siz[v]…

题目链接: http://codeforces.com/contest/629/problem/C 题意: 长度为n的括号,已经知道的部分的长度为m,现在其前面和后面补充‘(',或')',使得其长度为n,且每个左括号都能找到右括号与它匹配. 题解: dp[i][j]表示长度为i,平衡度为j的合法括号序列的总数,这里平衡度定义是‘('比')'多多少个,或')'比’('多多少个. #include<iostream> #include<cstring> #include<cstd…

B. Far Relative’s Problem time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Famil Door wants to celebrate his birthday with his friends from Far Far Away. He has n friends and each of them c…

E. Famil Door and Roads 题目连接: http://www.codeforces.com/contest/629/problem/E Description Famil Door's City map looks like a tree (undirected connected acyclic graph) so other people call it Treeland. There are n intersections in the city connected b…

D. Babaei and Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/D Description As you know, every birthday party has a cake! This time, Babaei is going to prepare the very special birthday party's cake. Simple cake is a cylinder of som…

A. Far Relative's Birthday Cake 题目连接: http://www.codeforces.com/contest/629/problem/A Description Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to make his birthday cake weird! The cake is a…

居然补完了 组合 A - Far Relative’s Birthday Cake import java.util.*; import java.io.*; public class Main { public static void main(String[] args) { Scanner cin = new Scanner (new BufferedInputStream (System.in)); int n = cin.nextInt (); int[] col = new int[…

题意:做蛋糕,给出N个半径,和高的圆柱,要求后面的体积比前面大的可以堆在前一个的上面,求最大的体积和. 思路:首先离散化蛋糕体积,以蛋糕数量建树建树,每个节点维护最大值,也就是假如节点i放在最上层情况下的体积最大值dp[i].每次查询比蛋糕i小且最大体积的蛋糕,然后更新线段树.注意此题查询的技巧!!查询区间不变l,r,才能保证每次查到的是小且最大体积. #include<iostream> #include<string> #include<algorithm> #in…

题意:n个人,在规定时间范围内,找到最多有多少对男女能一起出面. 思路:ans=max(2*min(一天中有多少个人能出面)) #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #includ…

水题 #include<iostream> #include<string> #include<algorithm> #include<cstdlib> #include<cstdio> #include<set> #include<map> #include<vector> #include<cstring> #include<stack> #include<cmath>…

A. Far Relative's Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

A. Far Relative’s Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

A. Far Relative’s Birthday Cake time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Door's family is going celebrate Famil Doors's birthday party. They love Famil Door so they are planning to m…

A. Far Relative's Birthday Cake 题意: 求在同一行.同一列的巧克力对数. 分析: 水题~样例搞明白再下笔! 代码: #include<iostream> using namespace std; const int maxn = 105; char a[maxn][maxn]; int main (void) { int N;cin>>N; int cnt = 0, res = 0; for(int i = 0; i < N; i++){ cn…

Famil Door's City map looks like a tree (undirected connected acyclic graph) so other people call it Treeland. There are n intersections in the city connected by n - 1 bidirectional roads. There are m friends of Famil Door living in the city. The i-t…

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 =…