Little Vova studies programming to p. Vova is very smart and he can write every test for any mark, but he doesn't want to stand out from the crowd too much. If the sum of his marks for all tests exceeds value x, then his classmates notice how smart h…

题目链接:https://vjudge.net/contest/226823#problem/B Little Vova studies programming in an elite school. Vova and his classmates are supposed to write n progress tests, for each test they will get a mark from 1 to p. Vova is very smart and he can write e…

http://codeforces.com/problemset/problem/540/B School Marks Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u Submit Status Practice CodeForces 540B Description Little Vova studies programming in an elite school. Vova and…

题意:先给定5个数,n,  k, p, x, y.分别表示 一共有 n 个成绩,并且已经给定了 k 个,每门成绩 大于0 小于等于p,成绩总和小于等于x, 但中位数大于等于y.让你找出另外的n-k个成绩. 析:利用贪心算法,首先是只能小于等于 p,也就是成绩越小越好, 然后中位数还得大于等于y,所以我们放的成绩只有两个,一个是1,另一个是y,那么是最优的, 所以每次只要判定这个就好,在判定的时候,要先排序,再找中位数,再就是每次放两个,如果n-k不是偶数,那么就放一个数,再进行. 代码如下: #…

题目链接:http://codeforces.com/problemset/problem/540/B 题目意思:给出 k 个test的成绩,要凑剩下的 n-k个test的成绩,使得最终的n个test之和 <= x,且中位数 >= y.凑的时候   1 <= test的成绩 <= p 做得可辛苦了,泪~~~ 首先统计给出的 k 个test中有多少个是 >= y 的,然后从后往前补充,直到中位数是 k ,其余的数用 1 来填充. 无解结果需要考虑清楚.(1)n-k+sum_k…

B. School Marks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little Vova studies programming in an elite school. Vova and his classmates are supposed to writen progress tests, for each…

codeforces 704B - Ant Man 贪心 题意:n个点,每个点有5个值,每次从一个点跳到另一个点,向左跳:abs(b.x-a.x)+a.ll+b.rr 向右跳:abs(b.x-a.x)+a.lr+b.rl,遍历完所有的点,问你最后的花费是多少 思路:每次选一个点的时候,在当前确定的每个点比较一下,选最短的距离. 为什么可以贪心?应为答案唯一,那么路径必定是唯一的,每个点所在的位置也一定是最短的. #include <bits/stdc++.h> using namespace…

CodeForces - 50A Domino piling (贪心+递归) 题意分析 奇数*偶数=偶数,如果两个都为奇数,最小的奇数-1递归求解,知道两个数都为1,返回0. 代码 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <sstream> #include <set> #include <map…

题目链接:http://codeforces.com/contest/161/problem/B 题意: 有n个商品和k辆购物车,给出每个商品的价钱c和类别t(1表示凳子,2表示铅笔),如果一辆购物车中有凳子,那么这辆购物车中最便宜的那个物品的价格能减少50%,问你如何放这些物品才能使总价钱最少. 思路: 简单贪心,判断凳子数量是否大于等于k行. #include <bits/stdc++.h> using namespace std; typedef long long LL; typede…

Trading Business 题目连接: http://codeforces.com/problemset/problem/176/A Description To get money for a new aeonic blaster, ranger Qwerty decided to engage in trade for a while. He wants to buy some number of items (or probably not to buy anything at al…

Shopping 题目连接: http://codeforces.com/gym/100803/attachments Description Your friend will enjoy shopping. She will walk through a mall along a straight street, where N individual shops (numbered from 1 to N) are aligned at regular intervals. Each shop…

题目链接:Codeforces 486C Palindrome Transformation 题目大意:给定一个字符串,长度N.指针位置P,问说最少花多少步将字符串变成回文串. 解题思路:事实上仅仅要是对称位置不同样的.那么指针肯定要先移动到这里,改动字符仅仅须要考虑两种方向哪种更优即 可. 然后将全部须要到达的位置跳出来.贪心处理. #include <cstdio> #include <cstring> #include <cstdlib> #include <…

题目链接:https://codeforces.com/contest/1154/problem/D 题解: 贪心思路,没有太阳的时候,优先用可充电电池走,万不得已才用普通电池走.有太阳的时候,如果可充电电池能够充一格电,就用普通电池跑(让可充电池充电),否则就用可充电电池走. AC代码: #include<bits/stdc++.h> using namespace std; ; int n,a,b; bool s[maxn]; int main() { cin>>n>&g…

Tennis Championship 题目链接:http://codeforces.com/problemset/problem/735/C ——每天在线,欢迎留言谈论. 题目大意: 给你一个 n (2≤n≤10^18),代表一共有n位参加比赛的选手. 游戏规则: ①每次比赛,输的选手将离开赛场 ②相互比赛的选手 他们的获胜的次数相差不能超过1(获胜4次的选手只能跟3或5次的选手比赛) 问题:最终赢得比赛的选手,胜场最多能为多少. 思路: 贪心:①选一名选手让他一直获胜且优先让他参加比赛 ②当…

题目:CodeForces - 363D 题意:给定n个学生,其中每个学生都有各自的私己钱,并且自己的私己钱只能用在自己买自行车,不能给别人. 给定m个自行车,每个自行车都有一个价格. 给定公有财产a.    然后求出这些学生能买自行车的最大数量,并且求当买下最大自行车数量时,总体花费私己钱的最少的钱. 我先来说以下二分搜索模板: //右值点不能取到的情况 int binary_search(vector<int>& nums,int left,int right, int targe…

Treeland is a country in which there are n towns connected by n - 1 two-way road such that it's possible to get from any town to any other town. In Treeland there are 2k universities which are located in different towns. Recently, the president signe…

Something happened in Uzhlyandia again... There are riots on the streets... Famous Uzhlyandian superheroes Shean the Sheep and Stas the Giraffe were called in order to save the situation. Upon the arriving, they found that citizens are worried about…

题目链接:http://codeforces.com/problemset/problem/777/E 题意:有n个环给你内环半径.外环半径和高度,叠这些环还要满足以下要求: ①:下面的环的外径要>=上面的环 ②:环不能掉下去,所以下面的环的内径要<上面的环的外径 ③:叠出最大高度 思路:贪心思想,先把这些环按外径从大到小,内径从大到小排好序,内径从大到小是为了尽可能让后面的环能放上去.接下来就可以用栈模拟了,放1~n个环的时每次都要判断,如果当前最顶上的环的内径大于我要放的环的内径也就是放不…

Wilbur and Array Time Limit: 2000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description Wilbur the pig is tinkering with arrays again. He has the array a1, a2, ..., an initially consisting of n zeros. At one step, he c…

链接 Codeforces 701E Connecting Universities 题意 n个点的树,给你2*K个点,分成K对,使得两两之间的距离和最大 思路 贪心,思路挺巧妙的.首先dfs一遍记录每个点的子树中(包括自己)有多少点是这K个中间的,第二遍dfs时对于每一条边,取两端包含较少的值,这样就保证树中间的点不会被取到,留下的就是相隔更远的点了.方法确实想不到啊. 代码 #include <iostream> #include <cstdio> #include <v…

题目描述: C. Elections time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output As you know, majority of students and teachers of Summer Informatics School live in Berland for the most part of the year…

原题链接:http://codeforces.com/contest/830/problem/A 题意:在一条数轴上分别有n个人和k把钥匙(n<=k),以及一个目的地,每个人要各自拿到一个钥匙后到达目的地.每个人的移动速度都是1, 问所有人都到达目的地的最短时间. 思路:转化一下题意,就是求耗时最长的人所用的最短时间. 我们可以二分答案x,然后对排序后的人以及钥匙进行枚举,进行从左至右搭配. 这里check函数中返回false的条件是从左至右所有人都能在x的时间内到达目的地,而计算这些人到达目的…

https://codeforces.com/problemset/problem/998/B 简单贪心题 代码如下: #include <stdio.h> #include <string.h> #include <iostream> #include <string> #include <math.h> #include <algorithm> #include <vector> #include <stack&…

Codeforces 题目传送门 & 洛谷题目传送门 一道不算太难的贪心,可惜又没自己想出来,显然省选之后我的能力呈 \(y=-1145141919810192608179998244353x+c\) 的趋势下滑,其中 \(c\) 为我省选前的能力( 首先假设我们已经选出了这 \(k\) 个操作,考虑按照怎样的顺序执行这些操作,显然我们会先执行赋值操作,再加法操作,最后乘法操作,因为赋值操作早晚都是要进行的,按照贪心的思想显然在一开始就进行赋值操作最优,而对于序列中某个元素 \(x\),我们假设…

题目链接:codeforces 725D . Contest Balloons 先按气球数从大到小排序求出初始名次,并把名次排在第一队前面的队放入优先队列,按w-t-1值从小到大优先,然后依次给气球给排名在前面的的队,给完后自己的气球数减少,继续跟之前排在第一队后面的队比较,考虑是否加入队列,每次记录最好的名次. #include<cstdio> #include<cstring> #include<algorithm> #include<queue> us…

题目链接: B. Passwords time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Vanya is managed to enter his favourite site Codehorses. Vanya uses n distinct passwords for sites at all, however he can…

题目链接: A. Exams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Student Valera is an undergraduate student at the University. His end of term exams are approaching and he is to pass exactly …

题目链接: A. Bits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's denote as  the number of bits set ('1' bits) in the binary representation of the non-negative integer x. You are given mu…

题意: 给n个奖池,t张彩票,q次操作. 每个奖池的奖金为pi. 每个奖池现有的彩票的数量为ai,保证ai>=1: q次操作,每次有两种,第i个奖池的现有彩票数量加一,或减一. 不允许投票的数量多于奖池数量的二分之一. 保证: n,t,q<=2e5 ai<=1000 pi<=1000 求在采用最佳策略的前提下获得奖金的期望. 思路: 首先要证明贪心的正确性,即把某张票投入某奖池之后其下一张票给期望做出的贡献要小于上一张彩票... 把式子写一下,求导,发现导数是单调递减的... 然后…

贪心,注意优先级,4单独,3与1先匹配,2与2匹配(注意判断2有没有剩下),然后2与两个1匹配,最后4个1匹配就可以了. #include<iostream> #include<cstdio> #include<cstring> using namespace std; int main() { int n; ],num; while(~scanf("%d",&n)) { memset(tot,,sizeof(tot)); ;i < n…