A. Bear and Big Brother time limit per test:1 second memory limit per test:256 megabytes input:standard input output:standard output Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak…

题目链接:A. Sereja and Swaps 题意:给定一个序列,能够交换k次,问交换完后的子序列最大值的最大值是多少 思路:暴力枚举每一个区间,然后每一个区间[l,r]之内的值先存在优先队列内,然后找区间外假设有更大的值就替换掉. 求出每一个区间的最大值,最后记录下全部区间的最大值 代码: By lab104_yifan, contest: Codeforces Round #243 (Div. 2), problem: (C) Sereja and Swaps, Accepted, #…

就暴力枚举所有起点和终点就行了. 我做这题时想的太多了,最简单的暴力枚举起始点却没想到...应该先想最简单的方法,层层深入. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<map> #include<set> #include<vector> #include<…

...不行.这题之后.不做1000分以下的了.很耻辱   A. Bear and Big Brother time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Bear Limak wants to become the largest of bears, or at least to become larger than his b…

A. time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Rock... Paper! After Karen have found the deterministic winning (losing?) strategy for rock-paper-scissors, her brother, Koyomi, comes up…

Bear Limak wants to become the largest of bears, or at least to become larger than his brother Bob. Right now, Limak and Bob weigh a and b respectively. It's guaranteed that Limak's weight is smaller than or equal to his brother's weight. Limak eats…

题目链接: http://codeforces.com/problemset/problem/653/C 题意: 给定序列,偶数位的数字比两边都大,则成为nice,只可以交换两个数字,问有多少种交换方式使得最后的序列为nice. 分析: 比赛的时候找规律,找呀找了好久都没看出来....由于只可以交换一次,交换两个数字,所以最多改变的6个数字的关系.那么可以交换成nice的话,原来不满足的肯定很少,直接把这些数字提出来,然后暴力的和序列每个元素交换一下,看看其余不满足的是否变成满足就好啦~~~ 代…

就枚举四种情况,哪种能行就是yes了.很简单,关键是写法,我写的又丑又长...看了zhanyl的写法顿时心生敬佩.写的干净利落,简直美如画...这是功力的体现! 以下是zhanyl的写法,转载在此以供学习: #include <vector> #include <list> #include <queue> #include <map> #include <set> #include <deque> #include <stac…

[题目链接]:click here~~ [题目大意] 一组题目的数目(n<=15),每一个题目有对应的难度,问你选择一定的题目(大于r个且小于l个)且选择后的题目里最小难度与最大难度差不小于x,求选择方案数. [解题思路]: DFS+回溯. 先发一发比較拙的代码: #include <bits/stdc++.h> using namespace std; const int N=1e5+10; int num[N],mum[N]; int n,m,q,t,l,r; int top,ans…

HDU 4930 Fighting the Landlords 题目链接 题意:就是题中那几种牌型.假设先手能一步走完.或者一步让后手无法管上,就赢 思路:先枚举出两个人全部可能的牌型的最大值.然后再去推断就可以 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct Player { int rank[15]; } p1, p2; int t,…

题目链接: http://www.codeforces.com/contest/666/problem/B 题意: 给你n个城市,m条单向边,求通过最短路径访问四个不同的点能获得的最大距离,答案输出一个满足条件的四个点. 题解: 首先预处理出任意两点的最短距离,用队列优化的spfa跑:O(n*n*logn) 现依次访问四个点:v1,v2,v3,v4 我们可以枚举v2,v3,然后求出v2的最远点v1,v3的最远点v4,为了保证这四个点的不同,直接用最远点会错,v1,v4相同时还要考虑次最远点来替换…

B. Covered Path Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/534/problem/B Description The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals v1 meters per…

题意:求定 n 个数,求有多少对数满足,ai^bi = x. 析:暴力枚举就行,n的复杂度. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <c…

A. Mike and palindrome time limit per test:2 seconds memory limit per test:256 megabytes input:standard input output:standard output Mike has a string s consisting of only lowercase English letters. He wants to change exactly one character from the s…

D. Diverse Garland time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have a garland consisting of nn lamps. Each lamp is colored red, green or blue. The color of the ii-th lamp is sisi ('…

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…

B. Laurenty and Shoptime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputA little boy Laurenty has been playing his favourite game Nota for quite a while and is now very hungry. The boy wants to make sa…

"Cricket Field" Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100002 Description Once upon a time there was a greedy King who ordered his chief Architect to build a field for royal cricket inside his park. The King was so…

题目链接: http://codeforces.com/gym/101194/attachments https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5930 题意: 现有四支队伍两两打比赛,总共就是打六场比赛,每场比赛赢的队伍可得 $3$ 分,输的队伍得 $0$ 分,平局则两个队各得 $1$ 分. 现在给出四个队伍最终的积分…

B. Lorry 题目连接: http://www.codeforces.com/contest/3/problem/B Description A group of tourists is going to kayak and catamaran tour. A rented lorry has arrived to the boat depot to take kayaks and catamarans to the point of departure. It's known that a…

http://codeforces.com/problemset/problem/633/D D. Fibonacci-ish   Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if the sequence consists of at least two elements f0 and f1 are ar…

[Codeforces 639F] Bear and Chemistry(Tarjan+虚树) 题面 给出一个n个点,m条边的无向图(不保证连通,可能有自环和重边),有q次询问,每次询问给出p个点和q条边,判断加上q条边后,这p个点中的任意一个点对(x,y)是否都满足:能从x走到y,再从y走到x,不经过重复的边.询问强制在线. \((1 ≤ n, q≤ 300 000, 0 ≤ m ≤ 300 000,\sum p \leq 300\ 000,\sum q \leq 300\ 000)\) 分析…

A. Hongcow Learns the Cyclic Shift time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Hongcow is learning to spell! One day, his teacher gives him a word that he needs to learn to spell. Bein…

/* 将给定的一个字符串分解成ABABA 或者 ABABCAB的形式! 思路:暴力枚举A, B, C串! */ 1 #include<iostream> #include<cstring> #include<cstdio> #include<string> using namespace std; string str; ]; int main(){ int t; scanf("%d", &t); getchar(); while…

题目链接:http://acm.hnu.cn/online/?action=problem&type=show&id=12886&courseid=274 解题报告:输入4个数,要你判断用 + .- . * ./.四种运算能不能得到一个结果为24的式子,可以用括号. 解释一下测试的第四组样例:应该是6 / (1 - 3 / 4) 暴力枚举三种符号分别是什么,然后枚举这三种符号运算的顺序,然后枚举这四个数字的24种排列方式,时间是4^3 * 6 * 24 然后注意要用double型,…

题目链接 题意:中文题. 题解:暴力枚举. #include <iostream> #include <cstring> using namespace std; ; ; char num[MAXS]; int main(int argc, const char * argv[]) { while (cin >> num) { ; int len = (int)strlen(num); ; ; i < len; i++) { ') { > MINK) { M…

昨天梦到这道题了,所以一定要A掉(其实梦到了3道,有两道记不清了) 暴力枚举等的是哪张牌,将是哪张牌,然后贪心的判断就行了. 对于一个状态判断是否为胡牌,1-n扫一遍,然后对于每个牌,先mod 3, 如果还有剩余,就需要和i+1,i+2凑成顺子,减掉i+1,i+2的值就行了,然后 只要枚举到的i为负就是不可行,还有要枚举到n+2位,因为第n,n-1为可能 减了n+1,n+2但是没扫到. /******************************************************…

http://poj.org/problem?id=3187 给定一个个数n和sum,让你求原始序列,如果有多个输出字典序最小的. 暴力枚举题,枚举生成的每一个全排列,符合即退出. dfs版: #include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <cstring> #include <string> #include <…

题意: 给出一个有n个点的无向图,每个点上有石头数个,现在的游戏规则是,设置某个点A的度数为d,如果A点的石子数大于等于d,则可以从A点给每个邻接点发一个石子.如果游戏可以玩10万次以上,输出INF,否则输出最多能玩几次. 思路: 暴力枚举每个可以玩的点,假如可以玩无限次,且当前状态为Z(指所有点的石头数的序列作为一个状态),那么在玩了多次之后,一定会造成循环,也就是说,玩几次之后,每个点的石子数和初始的石子数一模一样,这样子我再重复之前是怎么玩的就可以无限玩了.但是由于有200个点,所以玩一次…

问题描述 我们把一个数称为有趣的,当且仅当: 1. 它的数字只包含0, 1, 2, 3,且这四个数字都出现过至少一次. 2. 所有的0都出现在所有的1之前,而所有的2都出现在所有的3之前. 3. 最高位数字不为0. 因此,符合我们定义的最小的有趣的数是2013.除此以外,4位的有趣的数还有两个:2031和2301. 请计算恰好有n位的有趣的数的个数.由于答案可能非常大,只需要输出答案除以1000000007的余数. 输入格式 输入只有一行,包括恰好一个正整数n (4 ≤ n ≤ 1000). 输…