A. Warrior and Archer Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/594/problem/A Description In the official contest this problem has a different statement, for which jury's solution was working incorrectly, and for this…

C. Edo and Magnets Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/594/problem/C Description Edo has got a collection of n refrigerator magnets! He decided to buy a refrigerator and hang the magnets on the door. The shop ca…

D. Max and Bike Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/problem/D Description For months Maxim has been coming to work on his favorite bicycle. And quite recently he decided that he is ready to take part in a cy…

B. Pasha and Phone Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/problem/B Description Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of ex…

A. Vitaly and Night Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/595/problem/A Description One day Vitaly was going home late at night and wondering: how many people aren't sleeping at that moment? To estimate, Vitaly de…

题意: 这道英文题的题意稍稍有点复杂. 找长度为n的数字序列有多少种.这个序列可以分为n/k段,每段k个数字.k个数可以变成一个十进制的数Xi.要求对这每n/k个数,剔除Xi可被ai整除的情况,剔除X的第一个数(包括前导0)是bi的情况.问剩下的组合有多少种. 思路: 这题我是一波三折的.首先也没有考虑很多,看着可以暴力模拟过程,我就直接开始敲了,几个for循环敲出来,再把bug调一调和特殊情况考虑考虑,交了之后开始TLE,这时候意识到复杂度太大了,于是开始优化,做了(b[i])*(mmax/1…

题意:给你很多对数,要么是0要么是1.不全0则ans++. 思路即题意. #include<cstdio> #include<cstring> #include<iostream> using namespace std; #define MAX(x,y) (((x)>(y)) ? (x) : (y)) #define MIN(x,y) (((x) < (y)) ? (x) : (y)) #define ABS(x) ((x)>0?(x):-(x))…

C题题目出错了,unrating,2题就能有很好的名次,只能呵呵了. 水 A - Vitaly and Night /************************************************ * Author :Running_Time * Created Time :2015/11/8 星期日 22:41:11 * File Name :A.cpp ************************************************/ #include <b…

B. Pasha and Phone time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consis…

B. Pasha and Phone time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consis…

B. Pasha and Phone   Pasha has recently bought a new phone jPager and started adding his friends' phone numbers there. Each phone number consists of exactly n digits. Also Pasha has a number k and two sequences of length n / k (n is divisible by k) a…

D. Max and Bike For months Maxim has been coming to work on his favorite bicycle. And quite recently he decided that he is ready to take part in a cyclists' competitions. He knows that this year n competitions will take place. During the i-th competi…

Codeforces Round #633(Div.2) \(A.Filling\ Diamonds\) 答案就是构成的六边形数量+1 //#pragma GCC optimize("O3") //#pragma comment(linker, "/STACK:1024000000,1024000000") #include<bits/stdc++.h> using namespace std; function<void(void)> __…

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…

Codeforces Round #182 (Div. 1)题解 A题:Yaroslav and Sequence1 题意: 给你\(2*n+1\)个元素,你每次可以进行无数种操作,每次操作必须选择其中n个元素改变符号,你的目的是使得最后所有数的和尽量大,问你答案是多少 题解: 感觉上就是构造题,手动玩一玩就知道,当n为奇数的时候,你可以通过三次操作,使得只会改变一个负数的符号.同理n为偶数的时候,每次要改变两个负数的符号. 所以答案如下: 当n为奇数的时候,答案为所有数的绝对值和 当n为偶数的…