B. Ants Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/317/problem/B Description It has been noted that if some ants are put in the junctions of the graphene integer lattice then they will act in the following fashion: eve…

A. Even Odds 奇数个数\(\lfloor \frac{n+1}{2}\rfloor\) B. Strings of Power 从位置0开始,统计heavy个数,若当前为metal,则可以和之前的所有heavy配对. C. Perfect Pair 假设\(x\le y\),显然用\(x+y\)替换\(x\)可以达到最少步数. \(x\)会一直\(+y\)直到\(x'\gt y\). D. Ants 暴力模拟??? E. Balance 构出一棵树后,就随便做了. F. Game w…

B. ZgukistringZ Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/B Description Professor GukiZ doesn't accept string as they are. He likes to swap some letters in string to obtain a new one. GukiZ has strings a,…

A. PawnChess Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/592/problem/A Description Galois is one of the strongest chess players of Byteforces. He has even invented a new variant of chess, which he named «PawnChess». Thi…

A. Anton and Polyhedrons time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Anton's favourite geometric figures are regular polyhedrons. Note that there are five kinds of regular polyhedrons:…

A. Bus to Udayland time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder and Chris the Baboon are travelling to Udayland! To get there, they have to get on the special IOI bus. The…

B. Strings of Power Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/318/problem/C Description Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pair…

B. Strings of Power Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/318/problem/B Description Volodya likes listening to heavy metal and (occasionally) reading. No wonder Volodya is especially interested in texts concerning…

A. Even Odds Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/318/problem/A Description Being a nonconformist, Volodya is displeased with the current state of things, particularly with the order of natural numbers (natural n…

A. 2Char Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/593/problem/A Description Andrew often reads articles in his favorite magazine 2Char. The main feature of these articles is that each of them uses at most two distinc…

A. Straight «A» time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Noora is a student of one famous high school. It's her final year in school — she is going to study in university next year.…

B. Skills 题目连接: http://www.codeforces.com/contest/613/problem/B Description Lesha plays the recently published new version of the legendary game hacknet. In this version character skill mechanism was introduced. Now, each player character has exactly…

很简单的暴力枚举,却卡了我那么长时间,可见我的基本功不够扎实. 两个数相乘等于一个数6*n,那么我枚举其中一个乘数就行了,而且枚举到sqrt(6*n)就行了,这个是暴力法解题中很常用的性质. 这道题找出a和b中最小的那个,然后开始枚举,一直枚举到sqrt(6*n)的向上取整.这样所有可能是答案的情况都有啦.再干别的都是重复的或者肯定不是最小面积的. #include<iostream> #include<cstdio> #include<cstdlib> #includ…

将整个游戏可以划分成若干个互不相交的子游戏. 每个子游戏的sg值只与其中的数的个数有关.而这个数不会超过30. 于是可以预处理出这个sg值表. 然后从1到n枚举,对<=sqrt(n)的部分,用个set判重. 对于大于sqrt(n)的部分,统计其中不包含在之前已经划分出来的子游戏内的数的个数,如果是奇数,就再异或上1. /* #include<cstdio> #include<cstring> #include<set> #include<map> us…

题目传送门 /* 数论/暴力:找出第一次到a1,a2的次数,再找到完整周期p1,p2,然后以2*m为范围 t1,t2为各自起点开始“赛跑”,谁落后谁加一个周期,等到t1 == t2结束 详细解释:http://blog.csdn.net/u014357885/article/details/46044287 */ #include <cstdio> #include <algorithm> #include <cstring> #include <iostream…

题目传送门 /* 暴力:每次更新该行的num[],然后暴力找出最优解就可以了:) */ #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #include <string> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int num[MAXN];…

题目传送门 /* 暴力:O (n^2) */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> using namespace std; ; const int INF = 0x3f3f3f3f; int main(void) //Codeforces Round #183 (Div. 2) A. Pythago…

题目传送门 /* 数学/暴力:只要一个数的最后三位能被8整除,那么它就是答案:用到sprintf把数字转移成字符读入 */ #include <cstdio> #include <algorithm> #include <cstring> #include <iostream> #include <cmath> #include <vector> using namespace std; ; const int INF = 0x3f3…

题目传送门 /* 题意:问是否能用质量为w^0,w^1,...,w^100的砝码各1个称出重量m,砝码放左边或在右边 暴力/进制转换:假设可以称出,用w进制表示,每一位是0,1,w-1.w-1表示砝码与物品放在一起,模拟判断每位是否ok 详细解释:http://blog.csdn.net/u011265346/article/details/46556361 总结:比赛时压根没往进制去想,连样例也不知道是怎么回事..中文不行啊:( */ #include <cstdio> #include &…

题目传送门 /* 二分查找/暴力:先埃氏筛选预处理,然后暴力对于每一行每一列的不是素数的二分查找最近的素数,更新最小值 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int mn_r[MAXN]; int mn_c[MAXN]; bool is_prim…

题目传送门 /* 题意:删除若干行,使得n行字符串成递增排序 暴力+构造:从前往后枚举列,当之前的顺序已经正确时,之后就不用考虑了,这样删列最小 */ /************************************************ Author :Running_Time Created Time :2015-8-3 10:49:53 File Name :C.cpp *************************************************/ #in…

题目传送门 /* 构造+暴力:按照题目意思,只要10次加1就变回原来的数字,暴力枚举所有数字,string大法好! */ /************************************************ Author :Running_Time Created Time :2015-8-3 8:43:02 File Name :A.cpp *************************************************/ #include <cstdio>…

Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx  Solved: 2xx 题目连接 http://codeforces.com/contest/525/problem/D Description Finally it is a day when Arthur has enough money for buying an apartment. H…

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 #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 #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"…

题目大意 两个人轮流在一个字符串上删掉一个字符,没有字符可删的人输掉游戏 删字符的规则如下: 1. 每次从一个字符串中选取一个字符,它是一个长度至少为 3 的奇回文串的中心 2. 删掉该字符,同时,他选择的那个字符串分成了两个独立的字符串 现在问,先手是否必胜,如果先手必胜,输出第一步应该删掉第几个字符,有多解的话,输出序号最小的那个 字符串的长度不超过5000,只包含小写英文字母 做法分析 可以这样考虑:将所有的长度大于等于 3(其实只需要找长度为 3 的就行)的奇回文串的中心标记出来 我们将…

Codeforces Round #108 (Div. 2) C. Pocket Book 题意 给定\(N(N \le 100)\)个字符串,每个字符串长为\(M(M \le 100)\). 每次选择\(i, j, k\),然后交换串\(i\)和串\(j\)的长度为\(k\)的前缀. 操作可以做任意次,求最多能得到多少不同的字符串,\(modulo (10^9+7)\). 思路 相当于每个位置的可选字符为该列的不同字符的数量. 代码 C. Pocket Book D. Frames 题意 给定…

CF922 CodeForces Round #461(Div.2) 这场比赛很晚呀 果断滚去睡了 现在来做一下 A CF922 A 翻译: 一开始有一个初始版本的玩具 每次有两种操作: 放一个初始版本进去,额外得到一个初始版本和一个复制版本 放一个复制版本进去,额外得到两个复制版本 一开始有\(1\)个初始版本,是否能恰好得到\(x\)个复制版本和\(y\)个初始版本 Solution 傻逼题 要特判一些特殊情况(没有\(1A\)...) #include<iostream> #includ…

Codeforces Round #485 (Div. 2) https://codeforces.com/contest/987 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define IT set<ll>::iterator #define sqr(x) ((x)*(x)) #define pb…