PAT 1025 PAT Ranking(25 分)(排序+排名)

博客围绕PAT排名合并问题展开,介绍了编程能力测试PAT由浙江大学计算机学院组织,需编写程序合并各考点排名生成最终排名。给出了输入、输出规格及示例,题解提醒处理分数相同排名相同情况,并给出代码。

1025 PAT Ranking(25 分)

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

题解:注意处理分数相同排名相同的情况。

代码:

#include<bits/stdc++.h>
using namespace std;
struct Student{
    char id[15];
    int score;
    int location_number;  //考场号
    int local_rank;       //考场内排名
}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++;
        }
        sort(stu+num-k,stu+num,cmp);  //重点
        stu[num-k].local_rank=1;      //分数最高的排名第一
        for(int j=num-k+1;j<num;j++)
        {
            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;
}

 

### 关于PAT编程能力测试排名汇总 对于PAT编程能力测试中的排名汇总,在C语言版本的具体实现中,可以考虑如下方式来处理成绩统计与排名计算。通常情况下,这类问题涉及读取输入数据、解析这些数据并按照特定规则进行排序最后输出结果。 考虑到PAT平台可能存在的某些特性或限制,比如编辑器不支持`gets()`函数[^1],因此建议采用更安全可靠的输入方法如`scanf()`来进行字符串或其他类型的输入操作。 下面是一个简单的C语言代码片段用于模拟PAT排名汇总的功能: ```c #include <stdio.h> #include <stdlib.h> // 定义结构体存储参赛者信息 typedef struct { int id; float score; } Contestant; // 比较函数供qsort使用 int compare(const void *a, const void *b) { return ((Contestant *)b)->score - ((Contestant *)a)->score; } int main() { int N; // 参赛人数 scanf("%d", &N); Contestant contestants[N]; for (int i = 0; i < N; ++i) { contestants[i].id = i + 1; scanf("%f", &(contestants[i].score)); } qsort(contestants, N, sizeof(Contestant), compare); printf("Ranking:\n"); for (int i = 0; i < N; ++i) { printf("ID:%d Score:%.2f\n", contestants[i].id, contestants[i].score); } return 0; } ``` 此段代码实现了基本的成绩录入和基于数高低的排序功能,并打印出最终排名情况。需要注意的是实际比赛中可能会有更加复杂的要求,例如相同数如何处理等问题,则需进一步调整逻辑满足具体需求[^2]。
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