D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…

题意:给一篇文章,再给一些单词替换关系a b,表示单词a可被b替换,可多次替换,问最后把这篇文章替换后(或不替换)能达到的最小的'r'的个数是多少,如果'r'的个数相等,那么尽量是文章最短. 解法:易知单词间有二元关系,我们将每个二元关系建有向边,然后得出一张图,图中可能有强连通分量(环等),所以找出所有的强连通分量缩点,那个点的minR,Len赋为强连通分量中最小的minR,Len,然后重新建图,跑一个dfs即可得出每个强连通分量的minR,Len,最后O(n)扫一遍替换单词,统计即可. 代码…

After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3». The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered for…

After you had helped George and Alex to move in the dorm, they went to help their friend Fedor play a new computer game «Call of Soldiers 3». The game has (m + 1) players and n types of soldiers in total. Players «Call of Soldiers 3» are numbered for…

题目传送门 /* 题意:选择k个m长的区间,使得总和最大 01背包:dp[i][j] 表示在i的位置选或不选[i-m+1, i]这个区间,当它是第j个区间. 01背包思想,状态转移方程:dp[i][j] = max (dp[i-1][j], dp[i-m][j-1] + sum[i] - sum[i-m]); 在两个for循环,每一次dp[i][j]的值都要更新 */ #include <cstdio> #include <cstring> #include <algorit…

Codeforces Round #267 (Div. 2) C. George and Job题目链接请点击~ The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced the follow…

D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…

http://codeforces.com/contest/754/problem/D 题意: 给定几组区间,找k组区间,使得它们的公共交集最大. 思路: 在k组区间中,它们的公共交集=k组区间中右端点最小值-k组区间中左端点最大值.如果我们要区间大,那我们应该尽量让左端点小,右端点大. 先对区间按照左端点排序,然后用优先队列处理. 将区间按照左端点从小到大的顺序一一进队列,只需要进右端点即可,如果此时队列内已有k个数,则队首就是这k组区间的最小右端点,而因为左端点是从小到大的顺序进队列的,所以…

A #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set> using namespace std; #define N 100000 #def…

                                              C. George and Job   The new ITone 6 has been released recently and George got really keen to buy it. Unfortunately, he didn't have enough money, so George was going to work as a programmer. Now he faced t…

题目: A. George and Accommodation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output George has recently entered the BSUCP (Berland State University for Cool Programmers). George has a friend Al…

wa哭了,,t哭了,,还是看了题解... 8170436                 2014-10-11 06:41:51     njczy2010     C - George and Job             GNU C++     Accepted 109 ms 196172 KB 8170430                 2014-10-11 06:39:47     njczy2010     C - George and Job             GNU C…

题意:题目简化了就是要给你n个区间,然后让你选出k个区间  使得这k个区间有公共交集:问这个公共交集最大能是多少,并且输出所选的k个区间.如果有多组答案,则输出任意一种.   这题是用优先队列来处理区间问题的,感觉挺典型的所以记录下来.   首先,要知道 选取的k个区间的最大交集=区间右端点中的最小值-区间左端点中的最大值.那么,要求得这这么k个区间是公共交集最大,就创建一个最小堆的优先队列(只存放区间的右端点):然后按左端点从小到大(先将区间按左端点排序)将区间放入优先队列中.每当优先队列的大…

E. President and RoadsTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/567/problem/E Description Berland has n cities, the capital is located in city s, and the historic home town of the President is in city t (s ≠ t). The c…

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…

Codeforces Round #271 (Div. 2) A - Keyboard 题意 给你一个字符串,问你这个字符串在键盘的位置往左边挪一位,或者往右边挪一位字符,这个字符串是什么样子 题解 模拟一下就好了 代码 #include<bits/stdc++.h> using namespace std; string s[3]; map<char,int>r,c; char ss[2][107]; int main() { s[0]="qwertyuiop"…

Codeforces Round #177 (Div. 1) A. Polo the Penguin and Strings 题意 让你构造一个长度为n的串,且里面恰好包含k个不同字符,让你构造的字符串字典序最小. 题解 先abababab,然后再把k个不同字符输出,那么这样就是最少 代码 #include<bits/stdc++.h> using namespace std; string s; int main() { int n,k; scanf("%d%d",&am…