PAT 1012 The Best Rank 排序
To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C – C Programming Language, M – Mathematics (Calculus or Linear Algebra), and E – English. At the mean time, we encourage students by emphasizing on(强调) their best ranks — that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.
For example, The grades of C, M, E and A – Average of 4 students are given as the following:
Student ID C M E A
310101 98 85 88 90
310102 70 95 88 84
310103 82 87 94 88
310104 91 91 91 91
Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.
Input
Each input file contains one test case. Each case starts with a line containing 2 numbers N and M (<=2000), which are the total number of students, and the number of students who would check their ranks, respectively. Then N lines follow, each contains a student ID which is a string of 6 digits, followed by the three integer grades (in the range of [0, 100]) of that student in the order of C, M and E. Then there are M lines, each containing a student ID.
Output
For each of the M students, print in one line the best rank for him/her, and the symbol of the corresponding rank, separated by a space.
The priorities of the ranking methods are ordered as A > C > M > E. Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.
If a student is not on the grading list, simply output “N/A”.
Sample Input
5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999
Sample Output
1 C
1 M
1 E
1 A
3 A
N/A
题目意思:已知n个学生的3门科目的分数,平均分可以通过3门成绩计算得到,这样就会有4种排名C(C Programming Language)、M(Mathematics)、E(English)、A(Average),对于K次查询,给一个学生的ID,输出该学生在4种排名中最好的排名和排名方式,如果在多种排名方式中的排名相同,那么按照A>C>M>E的顺序输出,如果ID不存在
输出N/A。
解题思路:这个题已经有点成绩查询系统的意思了,核心是排序,需要注意的一点是排名并列应该是1、1、3、4、5而不是1、1、2、3、4。
#include<cstdio> #include<string> #include<iostream> #include<algorithm> using namespace std; struct student{ int id; int best; int score[4]; int ranks[4]; }stu[2010]; int vis[1000010]; int flag; int cmp(student A,student B) { return A.score[flag]>B.score[flag]; } int main() { int n,m,id,i,j,best,temp; char c[5]={'A','C','M','E'}; scanf("%d%d",&n,&m); for(i=0;i<n;i++) { scanf("%d%d%d%d",&stu[i].id,&stu[i].score[1],&stu[i].score[2],&stu[i].score[3]); stu[i].score[0]=(stu[i].score[1]+stu[i].score[2]+stu[i].score[3])/3.0;//平均分 } for(flag=0;flag<=3;flag++)//确定排名,保存在不同flag下的排名 { sort(stu,stu+n,cmp); stu[0].ranks[flag]=1; for(i=1;i<n;i++) { stu[i].ranks[flag]=i+1; if(stu[i].score[flag]==stu[i-1].score[flag])//成绩相同。排名将会并列 { stu[i].ranks[flag]=stu[i-1].ranks[flag]; } } } for(i=0;i<n;i++)//确定每个人在4种排名方式下的最好排名 { vis[stu[i].id]=i+1;//序号 stu[i].best=0; int mins=stu[i].ranks[0]; for(j=1;j<=3;j++) { if(stu[i].ranks[j]<mins) { mins=stu[i].ranks[j]; stu[i].best=j; } } } for(i=0;i<m;i++) { scanf("%d",&id); temp=vis[id]; if(temp) { best=stu[temp-1].best; printf("%d %c\n", stu[temp-1].ranks[best], c[best]); } else { printf("N/A\n"); } } return 0; }
原文地址:https://www.cnblogs.com/wkfvawl/p/11388397.html
- JavaScript 教程
- JavaScript 编辑工具
- JavaScript 与HTML
- JavaScript 与Java
- JavaScript 数据结构
- JavaScript 基本数据类型
- JavaScript 特殊数据类型
- JavaScript 运算符
- JavaScript typeof 运算符
- JavaScript 表达式
- JavaScript 类型转换
- JavaScript 基本语法
- JavaScript 注释
- Javascript 基本处理流程
- Javascript 选择结构
- Javascript if 语句
- Javascript if 语句的嵌套
- Javascript switch 语句
- Javascript 循环结构
- Javascript 循环结构实例
- Javascript 跳转语句
- Javascript 控制语句总结
- Javascript 函数介绍
- Javascript 函数的定义
- Javascript 函数调用
- Javascript 几种特殊的函数
- JavaScript 内置函数简介
- Javascript eval() 函数
- Javascript isFinite() 函数
- Javascript isNaN() 函数
- parseInt() 与 parseFloat()
- escape() 与 unescape()
- Javascript 字符串介绍
- Javascript length属性
- javascript 字符串函数
- Javascript 日期对象简介
- Javascript 日期对象用途
- Date 对象属性和方法
- Javascript 数组是什么
- Javascript 创建数组
- Javascript 数组赋值与取值
- Javascript 数组属性和方法
- c++ 优先队列(priority_queue)的详细讲解用法
- HDU 1022(关于栈的详细解法)
- count_if函数的用法
- Codeforces Round #633 (Div. 2) A ~~C
- AtCoder Beginner Contest 162 A~~D
- P1036 选数
- P1028 数的计算
- P1598 垂直柱状图
- 递归解决全排列问题
- Codeforces Round #622 (Div. 2)A~~C1
- AtCoder Beginner Contest 156 A~~D
- AtCoder Beginner Contest 155
- Codeforces Round #620 (Div. 2) A~~D
- DFS+记忆化搜索 -- 简单练习
- AtCoder Beginner Contest 154