题目链接:https://ac.nowcoder.com/acm/contest/903/B 题意: 给你 q,n,p,求 q1+q2+...+qn 的和 模 p. 思路:一开始不会做,后面查了下发现有个等比数列求和的快速幂公式,附上链接https://www.cnblogs.com/yuiffy/p/3809176.html AC代码: #include<cstdio> #include<iostream> using namespace std; long long mod;

刚刚过去的周六(6月7号)是今年高考的第一天,同一时候也是GPCT杯大学生程序设计大赛颁奖的日子,以下我们用图文再回想一下本次大赛颁奖的过程. 评审过程的一些花絮<感谢各位评审这些天的付出!>: 感谢葡萄城给我的这次的锻炼的机会,让我学会了非常多,也明确了非常多!最重要的是感谢评委们!你们辛苦了! 感谢MTV-我在上嵌入式课-- 永远别忘了为你付出的人,成就永远不属于一个人.就是装也得怀有一颗谦卑感恩之心.版主大人受小弟一拜! 再次感谢品审大人,我们组还非常多要改进的,到时候一定要跟各路大神好

题目描述 > “舔狗舔狗,> 舔到最后,> 一无所有.” 有 n 只舔狗,每只舔狗的心中都有自己朝思暮想的一位. 每个人虽然受到了一万次拒绝,还毅然第一万零一次鼓起勇气. 作为一个不食人间烟火的算法设计师,你早已看破红尘.但是,人世间的苦难仍让你挂念.看到众生在单恋中苦苦坚持,你决定普度众生,给大家找到一个最好的结局,让一无所有的舔狗尽量地少,让每个人都尽量能和自己喜欢的或喜欢自己的人修成正果. 也就是说,你需要给这 n 只舔狗配对,对于舔狗 i,他可以和他朝思暮想的人 aiai 配对.

题目链接:https://ac.nowcoder.com/acm/contest/903/L 题意:给你 n * n的方阵,你可以从任意一个数字开始走,可以走上下左右四个方向,走过的数字会被拼合,拼合的数字没有前导0,问最小不能拼合出来的数字是多少? 思路:开始就认为是一道深搜题,后面自己想太多就没去试,醉了orz! 这题就是把每个数字作为起点开始搜索,把已经搜索过的数字进行标记,然后从0开始检索,输出第一个未出现的数字就是答案.搜索的话我代码是搜5位数的所有组合,事实上这道题搜索3位数的所有组

题目链接:https://ac.nowcoder.com/acm/contest/903/J 题意:给你 n 个舔狗和他喜欢的人,让你俩俩配对(只能和喜欢它的和它喜欢的),求剩下的单身狗数量. 思路:类似于拓扑排序,由入度最少的边开始配对,也就是被最少的舔狗喜欢的(甚至是没有).将已经配对的舔狗进行标记,更新入度后重新加入优先队列,最后用总数减去标记数就是答案了. 总结:一开始我的思路是对的呐,但是我太菜了,卡在没办法处理同时配对2个点和维护他们入度,看完别人的处理才发现自己是局限于找入度为0而

UVA12504 - Updating a Dictionary 给出两个字符串,以相同的格式表示原字典和更新后的字典.要求找出新字典和旧字典的不同,以规定的格式输出. 算法操作: (1)处理旧字典,将旧字典中的每对关键字及其价值从字典串中截取出来,压入容器中.用做新旧字典对比检索. (2)处理新字典,将新字典中的每对关键字及其价值从字典串中截取出来.对于每个关键字,在容器中检索相同的关键字.若检索不成功,该关键字是新字典新增的,处理存储到相关的串数组中.若检索成功,提取容器中该关键字的价值与新

Demacia of the Ancients Time Limit: 2 Seconds Memory Limit: 65536 KB There is a popular multiplayer online battle arena game called Demacia of the Ancients. There are lots of professional teams playing this game. A team will be approved as Level K if

Capture the Flag Time Limit: 2 Seconds Memory Limit: 65536 KB Special Judge In computer security, Capture the Flag (CTF) is a computer security competition. CTF contests are usually designed to serve as an educational exercise to give participants ex

May Day Holiday Time Limit: 2 Seconds Memory Limit: 65536 KB As a university advocating self-learning and work-rest balance, Marjar University has so many days of rest, including holidays and weekends. Each weekend, which consists of Saturday and Sun

Lunch Time Time Limit: 2 Seconds Memory Limit: 65536 KB The 999th Zhejiang Provincial Collegiate Programming Contest will be held in Marjar University. The canteen of Marjar University is making preparations for this grand competition. The canteen pr

Beauty of Array Time Limit: 2 Seconds Memory Limit: 65536 KB Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all

Ace of Aces Time Limit: 2 Seconds Memory Limit: 65536 KB There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided to elect an outstandin

Team Formation Time Limit: 3 Seconds Memory Limit: 131072 KB For an upcoming programming contest, Edward, the headmaster of Marjar University, is forming a two-man team from N students of his university. Edward knows the skill level of each student.

Biggest Number http://acm.hust.edu.cn/vjudge/contest/view.action?cid=30851#problem/F 解题思路:DFS(检索)+BFS(探路)=强剪枝 一看就知道是搜索题,只是,要是DFS裸搜,真的会超时.时间,非常紧! 做该题的时候,要无时无刻不忘剪枝,抓住一切剪枝的机会,剪吧! (1)用flag记录当前检索的值与当前已经检索到的最大值的关系,一但有机会,就要剪掉那些检索到的值会比当前最大值要小的“残枝败叶”. (2)每次DF

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=30746#problem/D D - 平方根大搜索 UVA12505 - Searching in sqrt(n) 解题思路:求出n的平方根,去整数,化二进制,字符串子串查找. 基本思路简单,但是求平方根那里,不能直接用库函数方法,因为Math.sqrt()返回值的精度比较小.我们这里要用到的精度最高是140位.所以,求平方根的函数(中心代码转自http://blog.csdn.net/

problem 很妙啊--发现状态转移矩阵每一行都可以由上一行平移得到,每次只算第一行然后平移,\(O(n^3)\) 就变成了 \(O(n^2)\). #include <iostream> #include <cstdio> using namespace std; int n, k; struct Matrix{ int num[205][205]; Matrix operator*(const Matrix &x)const{ Matrix re; for(int i

Valley Numer Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 311    Accepted Submission(s): 165 Problem Description 众所周知,度度熊非常喜欢数字. 它最近发明了一种新的数字:Valley Number,像山谷一样的数字. 当一个数字,从左到右依次看过去数字没有出现先递增接

Pokémon GO Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 171    Accepted Submission(s): 104 Problem Description 众所周知,度度熊最近沉迷于 Pokémon GO. 今天它决定要抓住所有的精灵球! 为了不让度度熊失望,精灵球已经被事先放置在一个2*N的格子上,每一个格子上都

Arithmetic of Bomb Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 129    Accepted Submission(s): 94 Problem Description 众所周知,度度熊非常喜欢数字. 它最近在学习小学算术,第一次发现这个世界上居然存在两位数,三位数……甚至N位数! 但是这回的算术题可并不简单,由于

题目链接 :http://acm.hdu.edu.cn/showproblem.php?pid=6670 Mindis Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 548 Accepted Submission(s): 119 Problem Description 平面上有 n 个矩形,矩形的边平行于坐标轴,现在度度熊需要操控一名角