Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

D. Memory and Scores   Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, both Memory and Lexa get some integer in the range [ - k;k] (…

function script_faster_rcnn_demo() close all; clc; clear mex; clear is_valid_handle; % to clear init_key run(fullfile(fileparts(fileparts(mfilename('fullpath'))), 'startup')); %% -------------------- CONFIG -------------------- opts.caffe_version = '…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

做到这题时卡了不少时间,参考了别人的解法,觉得挺不错的,还挺巧妙. SELECT s2.Score,s1.Rank From ( SELECT S1.Score, COUNT(*) as Rank FROM (SELECT DISTINCT Score from Scores) as S1, (SELECT DISTINCT Score from Scores) as S2 Where S1.Score<=S2.Score Group By S1.Score ) s1 ,Scores s2 WH…

Using names.txt (right click and 'Save Link/Target As...'), a 46K text file containing over five-thousand first names, begin by sorting it into alphabetical order. Then working out the alphabetical value for each name, multiply this value by its alph…

Memory and Scores 题目链接:http://codeforces.com/contest/712/problem/D dp 因为每轮Memory和Lexa能取的都在[-k,k],也就是说每轮两人分数的变化量在[-2k,2k]: 故可以定义状态:dp[times][diff]为第times次Memory和Lexa的分数差为diff的方案数. 而dp[times][diff]可以从dp[times-1][diff-2k]到dp[times-1][diff+2k]转移而来: 又因为变化…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

Score 很好得到: select Score from Scores order by Score desc; 要得到rank, 可以通过比较比当前Score 大的Score 的个数得到: select Score, (select count(distinct Score) from Scores where Score>=s.Score) Rank where Scores s order by Score desc; 或者: select s.Score ,count(distinct…

Kyle is a student of Programming Monkey Elementary School. Just as others, he is deeply concerned with his grades. Last month, the school held an examination including five subjects, without any doubt, Kyle got a perfect score in every single subject…

D. Memory and Scores 题目连接: http://codeforces.com/contest/712/problem/D Description Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score b. In a single turn, b…

机器学习的评估 PR曲线用于positive类数据占比比较小,或者你更加在意false postion(相比于false negative):其他情况采用ROC曲线:比如Demo中手写体5的判断,因为只有少量5,所以从ROC上面来看分类效果不错,但是从PR曲线可以看到分类器效果不佳.   y_scores = sgd_clf.decision_function([some_digit]) decision_function代表的是参数实例到各个类所代表的超平面的距离:在梯度下滑里面特有的(随机森…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

题目链接    https://leetcode.com/problems/rank-scores/description/ 题意:对所有的分数按照降序进行排序,查询出分数和排名,排名相同的输出相同名次 此种解法在leetcode中未通过,看错误提示,好像是数据的精确度问题,不知道为什么. 附上一个通过的(mysql):https://blog.csdn.net/travel_1/article/details/51589706 思路: 1.首先查出 不重复的所有分数排名 select dist…

问题描述 解决方案 select sc.Score, (select count(*) from (select distinct Score from Scores ) ds where ds.Score>= sc.Score ) Rank from Scores sc order by Score desc…

# -*- coding: utf-8 -*- """ Created on Wed Aug 10 08:10:35 2016 @author: Administrator """ ''' 关于:cross_validation.scores 此处cross_validation.scores并不是cross_validation的scores, 而是分类函数(本文是clf,svm)的scores分成K部分K-1是训练1次是测试共K个scores…

#1236 : Scores 时间限制:4000ms 单点时限:4000ms 内存限制:256MB 描述 Kyle is a student of Programming Monkey Elementary School. Just as others, he is deeply concerned with his grades. Last month, the school held an examination including five subjects, without any do…

2015北京网络赛 J Scores 题意:50000组5维数据,50000个询问,问有多少组每一维都不大于询问的数据 思路:赛时没有思路,后来看解题报告也因为智商太低看了半天看不懂.bitset之前没用过,查了下发现其实就是一个二进制表示,这里的每一位就表示原序中的状态.  建一个bitset<50000> bs[6][sqrt(50000)], bs[i][j] 就表示在第i维上前j个块中的数据在原序中是哪些位置. #include <iostream> #include &l…

time limit per test2 seconds memory limit per test512 megabytes inputstandard input outputstandard output Memory and his friend Lexa are competing to get higher score in one popular computer game. Memory starts with score a and Lexa starts with score…

I believe that there are two types of people who get high scores in English exams: 1) have high intelligence. 2) spend time several times more than others on English learning so that their English levels outperform the level that the English exam req…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" be…

import java.util.*; public class ArraySortScore { //完成 main 方法 public static void main(String[] args) { int counter=0; int[] scores=new int[]{89 , -23 , 64 , 91 , 119 , 52 , 73}; SortScore(scores); int index=0; while(index<scores.length) { System.out…

Scores 求五维偏序(≤).强制在线. \(n,q\le 50000\) 烂大街的题解 考虑如果我们能用bitset找到每一维有哪些比询问点小,然后把每一维的结果取交集,得到的就是答案了. 然后对每一维分块处理前缀,询问时二分得到排名,将前面整块的和末尾的合并一下就行了. 时间复杂度\(O(\frac{nq}{32})\).分块对时间复杂度几乎没什么影响,重要的是bitset的操作. #include<bits/stdc++.h> #define co const #define il i…

[Codeforces712D] Memory and Scores(DP+前缀和优化)(不用单调队列) 题面 两个人玩游戏,共进行t轮,每人每轮从[-k,k]中选出一个数字,将其加到自己的总分中.已知两人的初始得分分别为a和b,求第一个人最后获胜的方案数.两种方案被认为是不同的,当且仅当存在其中一轮,其中一人选到的数字不同.a, b, t≤100,k≤1000 分析 两个人的操作是独立的,设\(dp1[i][j]\)表示第1个人玩i轮得到j分的方案数,第2个人同理 则有\(dp1[0][a]=…

目录 15.1 Outcome regression 15.2 Propensity scores 15.3 Propensity stratification and standardization 15.4 Propensity matching 15.5 Propensity models, structural models, predictive models Fine Point Nuisance parameters Effect modification and the prop…

题目大意: 两个人玩取数游戏,第一个人分数一开始是a,第二个分数一开始是b,接下来t轮,每轮两人都选择一个[-k,k]范围内的整数,加到自己的分数里,求有多少种情况使得t轮结束后a的分数比b高.  (1 ≤ a, b ≤ 100, 1 ≤ k ≤ 1000, 1 ≤ t ≤ 100) 1.我一开始的想法是DP出玩i轮得分是j的方案数.然后状态数最多有t*(2*k*t)那么多,最坏情况下会有2e7那么多的状态,转移必须是O(1)的. dp[i][j]=sum(dp[i-1][j-k....j+k]…

题目链接:http://codeforces.com/contest/712/problem/D A初始有一个分数a,B初始有一个分数b,有t轮比赛,每次比赛都可以取[-k, k]之间的数,问你最后A比B大的情况有多少种. dpA[i][j]表示第i轮获得j分的情况数.因为第i轮只和第i-1轮有关,所以这里i用滚动数组优化. 要是普通做法3个for就会超时,所以要用前缀和优化,dpA[i][j]可以由一段连续的dp[i - 1][x]转移过来,所以用sumA数组存取dp[i - 1][x]的前缀…