PAT A1025

最新推荐文章于 2024-09-03 22:15:15 发布

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本文介绍了一个程序设计任务，该任务旨在合并多个地点的Programming Ability Test (PAT)甲级竞赛成绩，并根据总分和注册号进行排名。通过使用结构体和自定义排序函数实现了对参赛者成绩的有效排序及排名。

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原题如下{ps.第一次接触PAT甲级}

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

代码如下：

其中利用了结构体，排序函数cmp

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct Student{
	char id[15];
	int score;
	int location_number;
	int local_rank;
}stu[30010];
bool cmp(Student a,Student b){
	if(a.score!=b.score) return a.score>b.score;
	else return strcmp(a.id,b.id)<0;
} 
int main(){
	int n,k,num=0;
	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++; 
	}
	printf("%d\t%d",k,num);
	sort(stu+num-k,stu+num,cmp);
	stu[num-k].local_rank=1;
	for(int j=num-k+1;j<num;j++){                    //这里不小心把int j=num-k+j,结果出来以后除了每组的第一名有localrank其他都为0
		if(stu[j].score==stu[j-1].score){
			stu[j].local_rank=stu[j-1].local_rank;
		}else{
			stu[j].local_rank=j+1-(num-k);
		}
	}
}
printf("%d\n",num);
sort(stu,stu+num,cmp);
int r=1;
for(int i=0;i<num;i++){
	if(i>0&&stu[i].score!=stu[i-1].score){
		r=i+1;
	}
	printf("%s ",stu[i].id);
	printf("%d %d %d\n",r,stu[i].location_number,stu[i].local_rank);
}
return 0;
}