C. Hacker, pack your bags!     It's well known that the best way to distract from something is to do one's favourite thing. Job is such a thing for Leha. So the hacker began to work hard in order to get rid of boredom. It means that Leha began to hac…

传送门 题意 给出n个区间[l,r]及花费\(cost_i\),找两个区间满足 1.区间和为指定值x 2.花费最小 分析 先用vector记录(l,r,cost)和(r,l,cost),按l排序,再设置一个数组bestcost[i]代表长度为i的最小花费. O(n)扫一遍,如果碰到区间左端点,更新答案:碰到右端点,更新bestcost[len],具体见代码 trick 1.更新答案会爆int 代码 #include <bits/stdc++.h> using namespace std; #d…

接上一篇文章; 这里直接把左端点和右端点映射到vector数组上; 映射一个open和close数组; 枚举1..2e5 如果open[i]内有安排; 则用那个安排和dp数组来更新答案; 更新答案完之后,如果有close数组 则把close数组里面的安排用来更新dp数组; #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #def…

Codeforces Round #422 (Div. 2) Table of Contents Codeforces Round #422 (Div. 2)Problem A. I'm bored with lifeProblem B.Crossword solvingProblem C. Hacker, pack your bags! Problem A. I'm bored with life A. I'm bored with life Holidays have finished. T…

[题目链接]:http://codeforces.com/contest/822/problem/C [题意] 有n个旅行计划, 每个旅行计划以开始日期li,结束日期ri,以及花费金钱costi描述; 让你在这n个旅行计划中选出两个计划; 要求这两个计划的日期没有相交的部分; 且这两个日期的总时间长度恰好为x; 让你求最小花费 [题解] 先把每个计划按照左端点第一优先级,右端点第二优先级升序排序; 然后定义一个dp[x]数组,表示在前i个计划中,时长为x,且右端点的位置< a[i].l的一个旅行…

E. Liar     The first semester ended. You know, after the end of the first semester the holidays begin. On holidays Noora decided to return to Vičkopolis. As a modest souvenir for Leha, she brought a sausage of length m from Pavlopolis. Everyone know…

B. Crossword solving     Erelong Leha was bored by calculating of the greatest common divisor of two factorials. Therefore he decided to solve some crosswords. It's well known that it is a very interesting occupation though it can be very difficult f…

A. I'm bored with life     Holidays have finished. Thanks to the help of the hacker Leha, Noora managed to enter the university of her dreams which is located in a town Pavlopolis. It's well known that universities provide students with dormitory for…

[题目链接]:http://codeforces.com/contest/822/problem/D [题意] 有n个人参加选美比赛; 要求把这n个人分成若干个相同大小的组; 每个组内的人数是相同的; 然后每个组内的人,两两比较; 每个组得出最美的人; 然后每个组中最美的人再重复上述步骤; 直到只剩一个人; 问你如何选定每个阶段的分组; 使得比较的次数最少; [题解] 只考虑一轮的情况; 设x是分组后每个组的人数; 然后一共有n个人; 则这一轮比较的次数就为 nx∗x∗(x−1)2 ->n∗(x…

[题目链接]:http://codeforces.com/contest/822/problem/B [题意] 让你用s去匹配t,问你最少需要修改s中的多少个字符; 才能在t中匹配到s; [题解] O(n2)的暴力搞就好; [Number Of WA] 1 [反思] 一开始判断的时候脑抽了; 写成只有s[1]==t[1]的时候才枚举; hack点是: 很多人两重for循环,没有给j层循环加限制; 直接两层循环1..n和1..m [完整代码] #include <bits/stdc++.h> u…

[题目链接]:http://codeforces.com/contest/822/problem/A [题意] 让你求a!和b!的gcd min(a,b)<=12 [题解] 哪个小就输出那个数的阶乘 [Number Of WA] 0 [反思] [完整代码] #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define L…

题意:给你两个串s,p,问你把s分开顺序不变,能不能用最多k段合成p. 题解:dp[i][j]表示s到了前i项,用了j段的最多能合成p的前缀是哪里,那么转移就是两种,\(dp[i+1][j]=dp[i][j],dp[i+lcp][j+1]=dp[i][j]+lcp\),这里的lcp是dp[i][j]和i的lcp,然后sa预处理一下st表就行了 //#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4)…

D. My pretty girl Noora     In Pavlopolis University where Noora studies it was decided to hold beauty contest "Miss Pavlopolis University". Let's describe the process of choosing the most beautiful girl in the university in more detail. The con…

传送门 题意 对于n个女孩,每次分成x人/组,每组比较次数为\(\frac{x(x+1)}{2}\),直到剩余1人 计算\[\sum_{i=l}^{r}t^{i-l}f(i)\],其中f(i)代表i个女孩的最少比较数 分析 难度在于如何计算f(i),f(i)每次除的是素数,详情见题解 那么我们对于每一个素数i,直接计算\(f[i]=\frac{x(x+1)}{2}\) 非素数,枚举能被i整除的第一个素数j,\(f[i]=f[i/j]+i*(j-1)/2\) -end- trick 代码 #inc…

今天老师(orz sansirowaltz)让我们做了很久之前的一场Codeforces Round #257 (Div. 1),这里给出A~C的题解,对应DIV2的C~E. A.Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular cho…

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"…