Jumping Jack CodeForces - 11B 就是一个贪心. 基本思路: 正负没有关系,先取绝对值. 首先跳过头,然后考虑怎么回来. 设超过头的步数为kk.如果kk为偶数,那么直接在前面跳过来的步数中选一个kk/2的步数,改成反着跳即可,不需要额外步数. 如果kk为奇数: 显然如果只把前面跳的改成反着跳不可能导致位置与目标位置差值的奇偶性变化,而kk为奇数,需要达到的差值0是偶数,因此一定需要额外步数. 那么可能有两种情况: 设当前已经跳了p步.如果p为偶数,那么这一步的下一步(p…

B. Jumping Jack time limit per test 1 second memory limit per test 64 megabytes input standard input output standard output Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get…

B. Jumping Jack 题目连接: http://www.codeforces.com/contest/11/problem/B Description Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has d…

B. Jumping Jack Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequen…

Jack is working on his jumping skills recently. Currently he's located at point zero of the number line. He would like to get to the point x. In order to train, he has decided that he'll first jump by only one unit, and each subsequent jump will be e…

题意: 给一个数X. 起始点为坐标0.第1步跳1格,第2步跳2格,第3步跳3格,.....以此类推. 每次可以向左跳或向右跳. 问最少跳几步可以到坐标X. 思路: 假设X是正数. 最快逼近X的方法是不停向右走.如果越过了X,假设到了X1,则必有X1-X小于最后一步d. 如果X1-X是偶数,将之前的某个x变为-x.则可以到X. 如果X1-X是奇数,因为将之前的某个x变为-x,实际是后退2x格.所以必定要再向右走一步.直到X1-X是偶数.(其实只要走一步或两步) X是负数的情况和X是正数的情况一样.…

A - Score UVA - 1585 水 #include<bits/stdc++.h> using namespace std; int main(){ int n; cin>>n; while(n--){ ; string s; cin>>s; int len=s.size(); ; ;i<len;i++){ if(s[i]=='O')sum+=tmp,tmp++; else { sum+=tmp; tmp=; } } sum+=tmp; cout<…

multi_match查询 multi_match查询提供了一个简便的方法用来对多个字段执行相同的查询. NOTE 存在几种类型的multi_match查询,其中的3种正好和在“了解你的数据”一节中提到的几种类型相同:best_fields,most_fields以及cross_fields. 默认情况下,该查询以best_fields类型执行,它会为每个字段生成一个match查询,然后将这些查询包含在一个dis_max查询中.下面的dis_max查询: { "dis_max": {…

  跨字段实体搜索  » 多数字段编辑 全文搜索被称作是 召回率(Recall) 与 精确率(Precision) 的战场: 召回率 ——返回所有的相关文档:精确率 ——不返回无关文档.目的是在结果的第一页中为用户呈现最为相关的文档. 为了提高召回率的效果,我们扩大搜索范围 ——不仅返回与用户搜索词精确匹配的文档,还会返回我们认为与查询相关的所有文档.如果一个用户搜索 “quick brown box” ,一个包含词语 fast foxes 的文档被认为是非常合理的返回结果. 如果包含词语 fa…

https://blog.csdn.net/Ricky110/article/details/78888711 多字段搜索多字符串查询boost 参数 “最佳” 值,较为简单的方式就是不断试错,比较合理的区间处于 1 到 10 之间,当然也有可能是 15 .如果为 boost 指定比这更高的值,将不会对最终的评分结果产生更大影响,因为评分是被 归一化的GET /_search{ "query": { "bool": { "should": [ {…