从以下这个三角形的顶部开始，向相邻的下一行的数字移动，经过之数所能得到的最大的和为23，即：\(3+7+4+9=23\) \[ 3\\ 7\quad4\\ 2\quad4\quad6\\ 8\quad5\quad9\quad3 \] 对于文本文件中包含的一百行的三角形(见数据文件ep67.txt)，求其从上到下所有路径中最大的和。 注：这个问题是第十八
Series.idxmax(self, axis=0, skipna=True, *args, **kwargs)[source] Return the row label of the maximum value. If multiple values equal the maximum, the first row label with that value is returned.
题目描述 Description Graph constructive problems are back! This time the graph you are asked to build should match the following properties. The graph is connected if and only if there exists a path between every pair of vertices. The diameter (aka "long
原题链接在这里：https://leetcode.com/problems/maximum-length-of-a-concatenated-string-with-unique-characters/ 题目： Given an array of strings arr. String s is a concatenation of a sub-sequence of arr which have unique characters. Return the maximum possib
What is the maximum number of MySQL servers in a group? A group can consist of maximum 9 servers. Attempting to add another server to a group with 9 members causes the request to join to be refused. This limit has been identified from testing and benchmar
原题链接在这里：https://leetcode.com/problems/path-with-maximum-gold/ 题目： In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty. Return the maximum amount of gold you can
洛咕 题意:给定\(n(n<=35)\)个数的数列和\(m(m<=1e9)\)，在数列中任选若干个数，使得他们的和对\(m\)取模后最大,求这个最大值. 分析:看到这个\(35\),很容易想到折半搜索.把数列分成相等的两个部分,分别爆搜出每个部分所有能够产生的和(在模\(m\)的意义下),时间复杂度为\(2^{18}\),很
1 typedef long long ll; 2 typedef pair<int,int> P; 3 #define _for(i,a,b) for(register int i = (a);i < b;i ++) 4 #define _rep(i,a,b) for(register int i = (a);i > b;i --) 5 #define INF 0x3f3f3f3f 6 #define MOD 100000000 7 #define maxn 10003 8
本文是“支持向量机系列”的第一篇，参见本系列的其他文章。 支持向量机即 Support Vector Machine，简称 SVM 。我最开始听说这头机器的名号的时候，一种神秘感就油然而生，似乎把 Support 这么一个具体的动作和 Vector 这么一个抽象的概念拼到一起，然后再做成一个 Machine ，一听就很玄了！
题目 考虑正难则反，答案即为\(n!-\text{返回值为n的排列数}\) 一个排列的返回值为\(n\)，当且仅当在\(n\)出现之前没有一个数后面有连续\(k\)个小于它的数 设\(f_i\)表示\(1\)到\(i\)的排列中，没有任何一个数后面有连续\(k\)个小于它的数 枚举\(i\)的位置，则有 \[f_i=\sum_{j=1}^kA_{i-
PyCUDA 可以通过 Python 访问 Navidia 的 CUDA 并行计算 API。 具体介绍和安装可以参考 PyCUDA 官网文档和 pycuda PyPI。 本文涵盖的内容有： 通过 PyCUDA 查询 GPU 信息。 NumPy array 和 gpuarray 之间的相互转换。 使用 gpuarray 进行基本的运算。 使用 ElementwiseKernel 进
ptrWaveBox.Axes(1).Maximum = 1000ptrWaveBox.Axes(2).Maximum = 20ptrWaveBox.Axes(2).Minimum = 0Dim fTemp(2000) As SingleFor n = 0 To iPixel - 1 fTemp(n) = wave(n)NextReDim wave(iPixel) As SingleFor n = 0 To iPixel - 1 wave(n) = fTemp(n)Nextwave(iPixe
class Solution { public int maxSubArray(int[] nums) { int ans = nums[0] ; int n = nums.length; int flag [] = new int[n]; flag[0] = nums[0]; for (int i = 1 ; i < n ; i++ ) { flag[i] = Math.max( fl
原题链接在这里：https://leetcode.com/problems/maximum-length-of-repeated-subarray/ 题目： Given two integer arrays A and B, return the maximum length of an subarray that appears in both arrays. Example 1: Input:A: [1,2,3,2,1]B: [3,2,1,4,7]Output: 3Explanation:
题目描述 思路分析 测试用例 Java代码 代码链接 题目描述 请定义一个队列并实现函数max得到队列里的最大值，要求函数max、push_back和 pop_front 的时间复杂度都是0(1)。 [牛客网刷题地址]无 思路分析 利用一个双端队列来存储当前队列里的最大值以及之后可能的
https://cn.vjudge.net/problem/POJ-3693 The repetition number of a string is defined as the maximum number R such that the string can be partitioned into R same consecutive substrings. For example, the repetition number of "ababab" is 3 and "
A tourist hiked along the mountain range. The hike lasted for n days, during each day the tourist noted height above the sea level. On the i-th day height was equal to some integer hi. The tourist pick smooth enough route for his hike, meaning that the be
#题目：使用lambda来创建匿名函数。 #!/usr/bin/python -- coding: UTF-8 -- MAXIMUM = lambda x,y : (x > y) * x + (x < y) * y MINIMUM = lambda x,y : (x > y) * y + (x < y) * x if name == ‘main’: a = 10 b = 20 print (‘The largar one is %d’ % MAXIMUM(a,b))
题目 Codeforces280D 原文 Consider integer sequence a1, a2, ..., an. You should run queries of two types: The query format is "0 i val". In reply to this query you should make the following assignment: ai = val. The query format is "1 l r k&quo
#include<stdio.h> #define MAX #define MAXIMUM(x,y)(x>y)?x:y #define MINIMUM(x,y)(x>y)?y:x int main() { int a=10,b=20; #ifdef MAX printf(“更大的数字是 %d\n”,MAXIMUM(a,b)); #else printf(“更小的数字是 %d\n”,MINIMUM(a,b)); #endif #ifndef MIN printf(“
题目链接 (BZOJ) https://www.lydsy.com/JudgeOnline/problem.php?id=3836 (Codeforces) http://codeforces.com/contest/280/problem/D 题解 似乎是最广为人知的模拟费用流题目。 线段树维护DP可以做，但是合并的复杂度是\(O(k^2)\), 会TLE. 考虑做\(k\)次费用流，很容易建出一个图
题意 给出n个不同的数字\(a_i\)，求出最大的子集，使得子集内任意两个数在二进制下至少有两位不同。 题解 先对任意两个二进制位只有一个不同的两个数连边，那么问题就转化成找出最多的点集，任意两点没有边，也就是最大独立集问题。普通的图求最大独立集是N-P困难的，但是二分图求最大独立集
You are given an array aa of nn integers, where nn is odd. You can make the following operation with it: Choose one of the elements of the array (for example aiai) and increase it by 11 (that is, replace it with ai+1ai+1). You want to make the me
报错详细信息：将索引进行online重建时，报错： SQL> alter index XX_DESC_INDEX rebuild online parallel 10; alter index XX_DESC_INDEX rebuild online tablespace XX_idx_new * ERROR at line 1: ORA-00604: error occurred at recursive SQL level 1 ORA-01450: maximum key l
A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach th
