C. The Monster time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As Will is stuck in the Upside Down, he can still communicate with his mom, Joyce, through the Christmas lights (he can turn t…

题意 给一个序列,包含(,),?,?可以被当做(或者),问你这个序列有多少合法的子序列. 分析 n^2枚举每一个子序列,暂时将每个?都当做右括号,在枚举右端点的时候同时记录两个信息:当前左括号多余多少个(top),已经将多少个?变成了右括号(cnt). 如果当前是(,top++. 如果当前是),top--. 如果当前是?,top--,cnt++; 如果top<0,我们需要判断一下当前还有没有之前被当做右括号的?,如果有的话,我们将其变为左括号,如果没有的话,意味着可以跳出循环了,之后都不会再合法…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 左括号看成1 右括号看成-1 设置l,r表示前i个数的和的上下界 遇到 左括号 l和r同时加1 遇到右括号 同时减1 遇到问号 因为问号可以为1或-1所以l减1而r加1 如果在某个时刻r小于0了 就说明再也不能继续了 因为说明左括号和问号比右括号来得少了 如果l<0则把l置为0,因为不允许和出现<0的情况 否则如果l等于0就说明有一种方法能让前i个数的和为0 当然当前枚举的序列长度要为偶数 [代码] #include <…

A. Eleven time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. El…

B. The Monster and the Squirrel Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/592/problem/B Description Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feeding her…

B. The Monster and the Squirrel time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Ari the monster always wakes up very early with the first ray of the sun and the first thing she does is feed…

D. MADMAX time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actu…

D. MADMAX time limit per test1 second memory limit per test256 megabytes Problem Description As we all know, Max is the best video game player among her friends. Her friends were so jealous of hers, that they created an actual game just to prove that…

B. Radio Station time limit per test2 seconds memory limit per test256 megabytes Problem Dsecription As the guys fried the radio station facilities, the school principal gave them tasks as a punishment. Dustin's task was to add comments to nginx conf…

C:显然可以设f[i][S]为当前考虑到第i位,[i,i+k)的状态为S的最小能量消耗,这样直接dp是O(nC(k,x))的.考虑矩阵快速幂,构造min+转移矩阵即可,每次转移到下一个特殊点然后暴力处理掉该点的贡献.可以预处理2p次转移矩阵进一步加速. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include&l…

扩欧,a+bx=c+dx,输出x>=0且y>=0,且a+bx最小的解. 要注意不能只保证x非负,还得看看能否保证y也非负. #include<cstdio> #include<iostream> using namespace std; typedef long long ll; ll a,b,c,d; void exgcd(ll a,ll b,ll &d,ll &x,ll &y) { if(!b) { d=a; x=1; y=0; } else…

A. Eleven time limit per test1 second memory limit per test256 megabytes Problem Description Eleven wants to choose a new name for herself. As a bunch of geeks, her friends suggested an algorithm to choose a name for her. Eleven wants her name to hav…

题意 在一个有向无环图上,两个人分别从一个点出发,两人轮流从当前点沿着某条边移动,要求经过的边权不小于上一轮对方经过的边权(ASCII码),如果一方不能移动,则判负.两人都采取最优策略,求两人分别从每个点出发的胜负关系表. 分析 记忆化搜索. f[x][y][v]表示现在两人分别在x,y,上一轮经过的边权为v时x是否能胜利(胜利为1,失败为0). 考虑如何转移: 对于一条从x到u的边权为val的边,如果val>=v,则可以走这条边,算出f[y][u][val], 如果f[y][u][val]为0…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用map模拟一下映射就好了. [代码] #include <bits/stdc++.h> using namespace std; int n,m; map<string,string> dic; int main(){ #ifdef LOCAL_DEFINE freopen("rush_in.txt", "r", stdin); #endif ios::sync_with_…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 这个数列增长很快的. 直接暴力模拟看看是不是它的一项就好了 [代码] #include <bits/stdc++.h> using namespace std; int n; bool ff(int x){ if (x==1) return true; int a = 1,b = 1,c=a+b; while (c<=x){ if (c==x){ return 1; } a = b,b = c; c = a+b; } r…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] f[x][y][z][2] 表示第一个人到了点x,第二个人到了点y,当前轮的字母(1..26),当前轮到谁走的情况下,谁赢. 写个记搜就好. 完全是模拟走的过程.. [代码] #include <bits/stdc++.h> using namespace std; const int N = 100; int n,m; vector <pair<int,int> > g[N+10]; int f[N+…

题意:给出一个串,只包含 ( ? ) 三种符号,求出有多少个子串是完美匹配的. ( ) ? ) => ( ) ( ) 完美匹配( ( ) ? => ( ( ) )完美匹配? ? ? ? => ( ) ( ) => ( ( ) ) 算一种子串 思路:看代码. #include<bits/stdc++.h> using namespace std; #define int long long signed main(){ string str; cin>>str…

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 =…