1025. PAT Ranking (25)解析

最新推荐文章于 2021-10-04 20:49:12 发布

原创最新推荐文章于 2021-10-04 20:49:12 发布 · 307 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#pat

本文介绍了一个用于合并多个考试地点的编程能力测试(PAT)排名列表的算法。该算法需要处理来自不同地点的输入数据，并根据特定规则进行排序，最终生成一个总体排名列表。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

PAT Ranking (25)
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
知识点：
头文件：#include

#include<algorithm>
using namespace std;
方式：sort(首元素地址（必填），尾元素地址的下一个地址（必填），比较函数（非必填））；

比较函数cmp：

#include <algorithm>
bool cmp(Student a,Student b){
if(a.score!=b.score) return a.score>b.score;//按分数从高到低
else return strcmp(a.name,b.name)<0;  //按名字从低到高排列
}

scanf("%s", char_type);//没有&

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct Student{
char id[13];  //考号
int score;  //分数
int location_number;
//考场号
int local_rank;
//考场内排名
int all_rank;  //总排名
}stu[30010];
bool cmp(Student a,Student b)
{ if(a.score!=b.score)
return a.score>b.score;
//分数从高到低排列 
    else return strcmp(a.id,b.id)<0; //id从低到高排列
}
int  main()
{
int n,k,num=0 ;
//n,考场数，k,考场人数,num，总考生数
scanf("%d",&n);
for (int i=1;i<=n;i++)
{scanf("%d",&k);
for (int j=0;j<k;j++)
{
scanf("%s %d",stu[num].id,&stu[num].score);
stu[num].location_number=i;
num++;

}
sort(stu+num-k,stu+num,cmp);
stu[num-k].local_rank=1;
for(int l=num-k+1;l<num;l++)
{
if(stu[l].score==stu[l-1].score)
stu[l].local_rank=stu[l-1].local_rank;
else stu[l].local_rank=l+1-(num-k);

}

}
sort(stu,stu+num,cmp);
stu[0].all_rank=1;
for(int j=1;j<num;j++)
{
if(stu[j].score==stu[j-1].score)
stu[j].all_rank=stu[j-1].all_rank;
else stu[j].all_rank=j+1;
}
printf("%d\n",num);
for(int m=0;m<num;m++)
{ printf("%s ",stu[m].id);
printf("%d %d %d\n",stu[m].all_rank,stu[m].location_number,stu[m].local_rank);
}
return 0;
}