A1025. PAT Ranking
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.
Output Specification:
For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:
registration_number final_rank location_number local_rank
The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.
Sample Input:
2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
Sample Output:
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
typedef struct{
char regist[];
int final_rank;
int location;
int local_rank;
int grade;
}info;
bool cmp(info a, info b){
if(a.grade != b.grade)
return a.grade > b.grade;
else
return strcmp(a.regist, b.regist) < ;
}
int main(){
int N, K, head = , rear = ;
info students[];
scanf("%d", &N);
for(int i = ; i < N; i++){
scanf("%d", &K);
head = rear;
for(int j = ; j < K; j++){
scanf("%s%d",students[rear].regist, &(students[rear].grade));
students[rear].location = i + ;
rear++;
}
sort(students + head, students + rear, cmp);
students[head].local_rank = ;
for(int j = ; j < K; j++){
if(students[head + j].grade == students[head + j - ].grade)
students[head + j].local_rank = students[head + j - ].local_rank;
else
students[head + j].local_rank = j + ;
}
}
sort(students, students + rear, cmp);
students[].final_rank = ;
printf("%d\n", rear);
printf("%s %d %d %d\n", students[].regist, students[].final_rank, students[].location, students[].local_rank);
for(int i = ; i < rear; i++){
if(students[i].grade == students[i - ].grade)
students[i].final_rank = students[i - ].final_rank;
else
students[i].final_rank = i + ;
printf("%s %d %d %d\n", students[i].regist, students[i].final_rank, students[i].location, students[i].local_rank);
}
cin >> N;
return ;
}
总结:
1、题目要求:按照最终排名的非降序排列,如果最终排名相同,则按照id的非降序排列。由样例可知,分数相同的人排名相同,但相同排名的人均会占一个位置(如有3个人并列第一, 则第四个人排名为第四而非第二)。
2、依旧使用struct记录学生成绩和信息。使用sort函数来排序。sort(首元素地址,尾元素的下一个地址,cmp函数),如排序a[0]~a[5],应填sort(a, a + 6, cmp)。在cmp函数中, 如果需要降序排序,则 return a > b,反之亦然。
3、字符串比较大小,可以使用strcmp(a, b),若a < b,返回负数,a = b返回0,a > b,返回正数。需要include <string.h>。
A1025. PAT Ranking的更多相关文章
- A1025 PAT Ranking (25)(25 分)
A1025 PAT Ranking (25)(25 分) Programming Ability Test (PAT) is organized by the College of Computer ...
- PAT A1025 PAT Ranking(25)
题目描述 Programming Ability Test (PAT) is organized by the College of Computer Science and Technology o ...
- PAT甲级——A1025 PAT Ranking
Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhe ...
- PAT甲级真题 A1025 PAT Ranking
题目概述:Programming Ability Test (PAT) is organized by the College of Computer Science and Technology o ...
- PAT A1025 pat ranking
有n个考场,每个考场都有若干数量个考生,现给出各个考场中考生的准考证号和分数,要求将所有考生的分数从高到低排序,并输出 #include<iostream> #include<str ...
- PAT_A1025#PAT Ranking
Source: PAT A1025 PAT Ranking Description: Programming Ability Test (PAT) is organized by the Colleg ...
- PAT Ranking (排名)
PAT Ranking (排名) Programming Ability Test (PAT) is organized by the College of Computer Science and ...
- 1025 PAT Ranking[排序][一般]
1025 PAT Ranking (25)(25 分) Programming Ability Test (PAT) is organized by the College of Computer S ...
- PAT 甲级 1025 PAT Ranking
1025. PAT Ranking (25) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Programmi ...
随机推荐
- js值----你所不知道的JavaScript系列(6)
1.数组 在 JavaScript 中,数组可以容纳任何类型的值,可以是字符串.数字.对象(object),甚至是其他数组(多维数组就是通过这种方式来实现的) .----<你所不知道的JavaS ...
- zookeeper Error contacting service 解决
连接kafka集群,有一个kafka机器连接失败 到该kafka机器上查询kafka进程,发现没有, 再查看zookeeper状态,提示 Error contacting service. It is ...
- CentOS6.9下升级默认的OpenSSH操作记录(升级到OpenSSH_7.6p1)
近期对IDC机房服务器做了一次安全漏洞扫描,漏扫结果显示服务器的OpenSSH版本太低(CentOS6默认是OpenSSH_5.3p1),存在漏洞隐患,安全部门建议升级到OpenSSH_7.6p1.升 ...
- 研究C语言的新型编译环境TCC
C语言综合研究1 搭建一个tcc环境 研究过程: 问题引出:为什么要使用tcc环境,甚至连图形界面都没有,为什么要使用这样的化境? 按照我们学习的本质来讲,可能是为了体验C语言底层的相关特性,但是在研 ...
- 《移山之道》Reading Task
老师布置的阅读任务虽然是附加的作业,但是对我来说是个很好的学习机会.软件工程主要是对工程的开发进行学习,毕竟在学校老师教了那么多的知识,我们课下做了那么多的练习并没有提高我们做一个工程的能力.一个项目 ...
- BugPhobia发布篇章:学霸在线系统测试报告
0x00 :测试报告版本管理 版本号 具体细节 修订时间 V 1.0 整理第一轮迭代用户管理和登陆注册的功能性验证测试,预计将继续网页对浏览器版本的兼容性测试 2015/11/12 V1.0.1 整理 ...
- Linux内核分析——程序破解
1. 掌握NOP.JNE.JE.JMP.CMP汇编指令的机器码 NOP:NOP指令即“空指令”.执行到NOP指令时,CPU什么也不做,仅仅当做一个指令执行过去并继续执行NOP后面的一条指令.(机器码: ...
- JavaScript —— 数组
Array方法 1.查找元素 indexOf()用来查找传进来的参数在目标数组中是否存在.如果目标数组包含该参数,就返回该元素在数组中的索引:如果不包含,就返回-1. 如果数组中包含多个相同的元素,i ...
- 可视化数据matplotlib之安装与简单折线图
matplotlib是一个可视化数据的模块,安装前需要先安装Visual Studio Community:然后去https://pypi.python.org/pypi上查找matplotlib并下 ...
- Python——多进程
进程的实例 # -*- coding:UTF-8 -*- import os import time from multiprocessing import Process #进程 def func( ...