zoukankan      html  css  js  c++  java
  • 1025 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 (≤), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (≤), 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

    题意:

      给出每个考生的成绩,求出其在考场中的排名和总排名。

    思路:

      模拟 + 排序

    Code:

     1 #include <bits/stdc++.h>
     2 
     3 using namespace std;
     4 
     5 struct Node {
     6     string registration_number;
     7     int final_rank;
     8     int location_number;
     9     int local_rank;
    10     int grade;
    11 };
    12 
    13 bool cmp(Node a, Node b) {
    14     if (a.grade == b.grade)
    15         return a.registration_number < b.registration_number;
    16     return a.grade > b.grade;
    17 }
    18 
    19 int main() {
    20     int n, k;
    21     cin >> n;
    22     string registration_number;
    23     int grade;
    24     vector<Node> testees;
    25     for (int i = 1; i <= n; ++i) {
    26         cin >> k;
    27         vector<Node> temp(k);
    28         for (int j = 0; j < k; ++j) {
    29             cin >> temp[j].registration_number >> temp[j].grade;
    30             temp[j].location_number = i;
    31         }
    32         sort(temp.begin(), temp.end(), cmp);
    33         int local_rank = 1;
    34         temp[0].local_rank = 1;
    35         testees.push_back(temp[0]);
    36         for (int j = 1; j < k; ++j) {
    37             local_rank++;
    38             if (temp[j].grade != temp[j - 1].grade)
    39                 temp[j].local_rank = local_rank;
    40             else
    41                 temp[j].local_rank = temp[j - 1].local_rank;
    42             testees.push_back(temp[j]);
    43         }
    44     }
    45     sort(testees.begin(), testees.end(), cmp);
    46     int final_rank = 1;
    47     testees[0].final_rank = 1;
    48     for (int i = 1; i < testees.size(); ++i) {
    49         final_rank++;
    50         if (testees[i].grade != testees[i - 1].grade)
    51             testees[i].final_rank = final_rank;
    52         else
    53             testees[i].final_rank = testees[i - 1].final_rank;
    54     }
    55     cout << testees.size() << endl;
    56     for (int i = 0; i < testees.size(); ++i) {
    57         cout << testees[i].registration_number << " " << testees[i].final_rank
    58              << " " << testees[i].location_number << " "
    59              << testees[i].local_rank << endl;
    60     }
    61     return 0;
    62 }
  • 相关阅读:
    对于bilibili主页head部分的代码的总结以及疑问。
    小项目使用框架(B/S),简单实现mvc分离
    petshop4学习_重构DataList实现分页
    照书抄了个小的sniffer。不过有问题测试不能通过,请求各位帮帮我
    《灵魂的迷茫》
    vs2005中gridview的RowCommand事件
    快速得到google的Gmail(已申请的免看)
    命令行下重启/启动停止IIS服务器的命令
    ES6之const命令
    不容忽视的js面试题
  • 原文地址:https://www.cnblogs.com/h-hkai/p/13157647.html
Copyright © 2011-2022 走看看