Song Jiang's rank list】的更多相关文章

 Song Jiang's rank list Time Limit:1000MS     Memory Limit:512000KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 5131 Description <Shui Hu Zhuan>,also <Water Margin>was written by Shi Nai'an -- an writer of Yuan and Ming dy…
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5131 Song Jiang's rank list Description <Shui Hu Zhuan>,also <Water Margin>was written by Shi Nai'an -- an writer of Yuan and Ming dynasty. <Shui Hu Zhuan>is one of the Four Great Classical…
欢迎参加——每周六晚的BestCoder(有米!) Song Jiang's rank list Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 673    Accepted Submission(s): 333 Problem Description <Shui Hu Zhuan>,also <Water Margin&g…
Song Jiang's rank list Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2006    Accepted Submission(s): 1128 Problem Description <Shui Hu Zhuan>,also <Water Margin>was written by Shi Nai…
Song Jiang's rank list Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)Total Submission(s): 2006    Accepted Submission(s): 1128 Problem Description <Shui Hu Zhuan>,also <Water Margin>was written by Shi Nai…
https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=5085 题目大意:给n个人以及每个人的名字s和他能杀死的敌人数m,将这些人按能杀死的敌人数进行排名并输出,m次查询,每次查询输出这个人的两个排名, 一个是他在他排名之前(他不能杀死,注:他不能杀死与他杀死敌人数相等的人)的人数 + 1, 一个是他排名之前与他杀死敌人数相等的人数…
STL的使用. /* 5131 */ #include <iostream> #include <map> #include <cstdio> #include <cstdlib> #include <cstring> #include <string> #include <algorithm> using namespace std; #define MAXN 205 typedef struct { string s;…
难点在于:递归函数和输出: #include <iostream> #include <vector> #include <algorithm> #include <map> using namespace std; class Rank{ private: int length,width,ans=0; char c; vector <char> v; int a[26][1000]; map <char,int> m; //out…
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…
排序: ---rank()over(order by 列名 排序)的结果是不连续的,如果有4个人,其中有3个是并列第1名,那么最后的排序结果结果如:1 1 1 4select scoreid, studentid,COURSENAME,totalexamscore ,rank()over(order by TOTALEXAMSCORE desc)orderbyNumfrom SCORECOURSE a ,COURSESCORE bwhere a.SCORECOURSEID = b.SCORECO…