CodeForces - 589B(暴力+排序)】的更多相关文章

Dasha decided to bake a big and tasty layer cake. In order to do that she went shopping and bought n rectangular cake layers. The length and the width of the i-th cake layer were ai and birespectively, while the height of each cake layer was equal to…
题意:给定 n 个矩形是a*b的,问你把每一块都分成一样的,然后全放一块,高度都是1,体积最大是多少. 析:这个题,当时并没有完全读懂题意,而且也不怎么会做,没想到就是一个暴力,先排序,先从大的开始选,如果大,那么数量少,如果小,数量就多, 用一个multiset来排序,这样时间复杂度会低一点,每一个都算一下比它的大矩阵的数量,然后算体积,不断更新,最大值. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #i…
题目链接:http://codeforces.com/problemset/problem/589/B 题目大意:告诉你n 个矩形,知道矩形的长度和宽度(长和宽可以互换),每个矩形的长度可以剪掉一部分,宽度也同样,求出可以得到多少个一样(每个矩形的长相等,宽也相等)的矩形,使得面积之和最大,并输出长和宽. 解题思路:将每个矩形的长和宽都存到一个结构体数组内,对矩形的长进行排序,因为最终结果的长一定是某个矩形的长,宽也一定是某个矩形的宽,所以我们枚举长,然后将长大于等于该长度的矩形的宽加入到一个向…
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传送门:https://codeforces.com/contest/691/problem/F 题意:给你n个数和q次询问,每次询问问你有多少对ai,aj满足ai*aj>=q[i],注意 a*b 与 b*a是不同的 题解:sum[i]记录的是两个数乘积为i的方法数,然后前缀和记录小于等于乘积为i的方法个数,输出答案就容斥一下,因为n个数最多组成n*(n-1)/2对数,减去小于乘积为q[i]的数后即为乘积大于等于q[i]的方法个数 为什么可以暴力是因为注意到了值域的范围为3e6,调和级数的复杂度…
题意:一个长度为n的字符串(只包含26个小字母)有q次操作 对于每次操作 给一个区间 和k k为1把该区间的字符不降序排序 k为0把该区间的字符不升序排序 求q次操作后所得字符串 思路: 该题数据规模很大 排序是关键想到计数排序,根据计数排序原理,由只有26个小写字母,需要统计区间字母的个数,还需要更新区间,想到用线段树优化,对于每个字母建一个线段树维护各字母在区间的个数. #include <map> #include <set> #include <list> #i…