题目链接: [传送门][1] Pasha Maximizes time limit per test:1 second     memory limit per test:256 megabytes Description Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he should make it larger. Unfortu…

简单贪心.... B. Pasha Maximizes time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pasha has a positive integer a without leading zeroes. Today he decided that the number is too small and he shoul…

题意:给出一串数字,给出k次交换,每次交换只能交换相邻的两个数,问最多经过k次交换,能够得到的最大的一串数字 从第一个数字往后找k个位置,找出最大的,往前面交换 有思路,可是没有写出代码来---sad #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack> #include<vector> #include<ma…

B. Pasha and Tea time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Pasha decided to invite his friends to a tea party. For that occasion, he has a large teapot with the capacity of w millilit…

题目链接: 传送门 Lawnmower time limit per test:2 second     memory limit per test:256 megabytes Description You have a garden consisting entirely of grass and weeds. Your garden is described by an n × m grid, with rows numbered 1 to n from top to bottom, an…

Soldier and Badges time limit per test 3 seconds memory limit per test 256 megabytes input standard input output standard output Colonel has n badges. He wants to give one badge to every of his n soldiers. Each badge has a coolness factor, which show…

Anya and Ghosts time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Anya loves to watch horror movies. In the best traditions of horror, she will be visited by m ghosts tonight. Anya has lots…

题面: 传送门 思路: 一眼看得,这是贪心[雾] 实际上,我们要求的答案就是sigma(ci*(ti-i))(i=1~n),这其中sigma(ci*i)是确定的 那么我们就要最小化sigma(ci*ti) 所以在新的每一秒,就把这一秒开始可以起飞的飞机中,cost最大的那一个拿出来,让他起飞就可以了 证明: 设最大的为m,我们取得另一个为n 那么n*ti+m*(ti+1) >= n*(ti+1)+m*ti 所以取m最好 这个过程用堆实现,懒得手打了,就用了priority_queue Code:…

题目链接:http://codeforces.com/problemset/problem/435/B 题目意思:给出一个最多为18位的数,可以通过对相邻两个数字进行交换,最多交换 k 次,问交换 k 次之后,这个数最大可以变成多少. 不知道最近是不是疏于训练(一直研究百度之星的题目,最终决定就是暂时放下,可能能力还没达到做那种题目的水平,不过都好感谢乌冬兄耐心甘为我解答左两道题目),昨晚又想学学拓扑排序(SPFA提到),结果没看明白= =...再加上昨晚比赛...电脑卡机卡得要命,于是悲催了=…

传送门 解题思路 贪心.对于一段区间中,可以将这段区间中相同的元素同时变成\(c\),但要付出的代价是区间中等于\(c\)的数的个数,设\(sum[i]\)表示等于\(c\)数字的前缀和,Max[i]表示数字\(i\)的最大个数.那么只要\(O(n)\)的扫一遍,维护一下每个数字的\(max\),具体做法是看一下\(Max[a[i]]\)大还是\(sum[i]\)大,如果\(sum\)大的话,说明前面都不变,直接把\(Max\)赋值成\(sum[i]+1\),否则直接让\(Max[i]++\),…