#include<bits/stdc++.h> using namespace std; int a[100001]; double ave[100001]; struct student{ int i; double ave; }; student stu[100001]; bool cmp(const student &s1,const student &s2){ if(s1.ave==s2.ave){ return s1.i<s2.i; }else{ return…

题目链接 题目要求: Given a list of non negative integers, arrange them such that they form the largest number. For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. Note: The result may be very large, so you need to return a string inst…

题目 Given a list of non negative integers, arrange them such that they form the largest number. For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. Note: The result may be very large, so you need to return a string instead of a…

题目:最大的矩形柱状图 难度:hard 题目内容: Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram. 翻译: 给定n个非负整数表示直方图的杆高度,其中每个条的宽度为1,找出直方图中最大矩形的面积. Above is a histogra…

Given a list of non negative integers, arrange them such that they form the largest number. For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. Note: The result may be very large, so you need to return a string instead of an i…

Largest Point Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Total Submission(s): 1102    Accepted Submission(s): 429 Problem Description Given the sequence A with n integers t1,t2,⋯,tn. Given the integral coeffic…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1506 刚开始没考虑时间复杂度,直接敲了,直接tle了,之后没有思路,然后看题解,看见大神写的优化非常棒. 大神的解释:(其实对于w[i]来说,如果去求后面连续的值,完全没必要一个个去比对,直接看w[i-1]的值就行了. 比如说2.3.4.5这个序列,如果我们要看3往后能延伸多长,并不需要去逐个和4和5比较,在计算4的时候,我们已经计算过5是比4大的,因为3比4小,4所能往后延伸的长度,3也一定能达到(4…

浅谈栈:https://www.cnblogs.com/AKMer/p/10278222.html 题目传送门:http://poj.org/problem?id=2559 贪心的想,最大的子矩阵顶部肯定会与某个矩阵的顶部重合,所以我们可以考虑对于每个矩阵,如果一个子矩阵的高度与它相同,那么向左向右分别可以延伸多长即可.用单调栈维护,从前往后从后往前扫两遍统计答案即可. 时间复杂度:\(O(n)\) 空间复杂度:\(O(n)\) 代码如下: #include <cstdio> #include…

题目传送门:https://agc009.contest.atcoder.jp/tasks/agc009_e 题目翻译 纸上写了\(N\)个\(1\)和\(M\)个\(0\),你每次可以选择\(k\)个数字擦掉,然后再写一个他们的平均值上去.保证\(N+M-1\)可以整除\(k-1\),请问最后留下来的那个数有多少种.\(N,M,K\leqslant 2000\) 题解 这题可以转化一下题意:有一棵\(k\)叉树,有\(n+m\)个叶子,每个叶子的权值是\(0\)或\(1\),其他结点的权值是子…

传送门 好神啊 直接考虑一棵 \(n+m\) 个叶子的 \(k\) 叉树,根结点权值为 \(\sum_{i\in m}(\frac{1}{k})^{deep_i}\) 对于一个 \(deep\) 的序列 如果 \(\sum_{i\in m}(\frac{1}{k})^{deep_i}+\sum_{i\in n}(\frac{1}{k})^{deep_i}=1\) 那么一定可以构造出一棵 \(k\) 叉树满足要求 (从deep大到小考虑,除去 \(k\) 进位) 那么对于一个答案数 \(x\),它…