• PAT A1025 PAT Ranking(25)


    题目描述

    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.
    

    输入格式

    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.
    

    输出格式

    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.
    

    输入样例

    2
    5
    1234567890001 95
    1234567890005 100
    1234567890003 95
    1234567890002 77
    1234567890004 85
    4
    1234567890013 65
    1234567890011 25
    1234567890014 100
    1234567890012 85
    

    输出样例

    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
    

    全部AC

    #include <cstdio>
    #include <iostream>
    #include <algorithm>
    #include <cstring>
    //p115
    using namespace std;
    
    const int maxn = 30010;
    
    struct Student {
        char registration_number[20]; //考生id
        int score; //考生成绩
        int final_rank; //最终排名
        int location_number; //考室号
        int local_rank; //考室排名
    }stu[maxn];
    
    bool cmp(Student A, Student B) {
        if(A.score != B.score) return A.score > B.score; //AB成绩不同,按成绩从大到小排序
        else if(A.score == B.score) return strcmp(A.registration_number, B.registration_number) < 0; //AB成绩相同,则按照准考证从小到大排序
    }
    
    int main(){
        #ifdef ONLINE_JUDGE
        #else
            freopen("1.txt", "r", stdin);
        #endif // ONLINE_JUDGE}
        int N; //考室数量
        scanf("%d", &N);
        int a = 0; //初始化考生存储位置
        for(int i = 0; i < N; i++) {
            int K; //考室中考生的数量
            scanf("%d", &K);
            //输入各个考室考生的注册号和成绩
            for(int j = 0; j < K; j++) {
                stu[a].location_number = i + 1;
                scanf("%s %d", stu[a].registration_number, &stu[a].score);
                a++;
            }
        }
        //初始化每个教室排名
        int lrank[N+1] = {0};
        int lrealrank[N+1] = {0};
        //memset(lrank, 1, N+1);
        //按考生成绩从大到小排序
        sort(stu, stu+a, cmp);
        int rank = 1; //总排名
        for(int i = 0; i < a; i++) {
            //printf("registration_number:%s location_number:%d score:%d 
    ", stu[i].registration_number, stu[i].location_number, stu[i].score);
            //总排名
            if(i == 0) stu[i].final_rank = rank;
            else if(i != 0 && stu[i].score == stu[i-1].score) stu[i].final_rank = stu[i-1].final_rank;
            else if(stu[i].score < stu[i-1].score) stu[i].final_rank = ++rank;
            rank = i+1; //更新总排名
            //教室排名
            if(i != 0 && stu[i].score == stu[i-1].score && stu[i].location_number == stu[i-1].location_number) { //在同个教室中分数相同
                    stu[i].local_rank = stu[i-1].local_rank;
                    lrealrank[stu[i].location_number]++;
            } else {
                stu[i].local_rank = lrank[stu[i].location_number]; //在同个教室中分数不相同
                lrank[stu[i].location_number]++;
                lrealrank[stu[i].location_number]++;
            }
            lrank[stu[i].location_number] = lrealrank[stu[i].location_number];
        }
        printf("%d
    ", a);
        for(int i = 0; i < a; i++) {
            printf("%s %d %d %d
    ", stu[i].registration_number, stu[i].final_rank, stu[i].location_number, stu[i].local_rank+1);
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/isChenJY/p/11444667.html
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