PAT B1015/A1062 Talent and Virtue

题目描述：

　　About 900 years ago, a Chinese philosopher Sima Guang wrote a history book in which he talked about people‘s talent and virtue. According to his theory, a man being outstanding in both talent and virtue must be a "sage（圣人）"; being less excellent but with one‘s virtue outweighs talent can be called a "nobleman（君子）"; being good in neither is a "fool man（愚人）"; yet a fool man is better than a "small man（小人）" who prefers talent than virtue.
Now given the grades of talent and virtue of a group of people, you are supposed to rank them according to Sima Guang‘s theory.

　　Input Specification:
　　Each input file contains one test case. Each case first gives 3 positive integers in a line: N (≤10?5??), the total number of people to be ranked; L (≥60), the lower bound of the qualified grades -- that is, only the ones whose grades of talent and virtue are both not below this line will be ranked; and H (<100), the higher line of qualification -- that is, those with both grades not below this line are considered as the "sages", and will be ranked in non-increasing order according to their total grades. Those with talent grades below H but virtue grades not are cosidered as the "noblemen", and are also ranked in non-increasing order according to their total grades, but they are listed after the "sages". Those with both grades below H, but with virtue not lower than talent are considered as the "fool men". They are ranked in the same way but after the "noblemen". The rest of people whose grades both pass the L line are ranked after the "fool men".
　　Then N lines follow, each gives the information of a person in the format:
　　ID_Number Virtue_Grade Talent_Grade
　　where ID_Number is an 8-digit number, and both grades are integers in [0, 100]. All the numbers are separated by a space.

　　Output Specification:
　　The first line of output must give M (≤N), the total number of people that are actually ranked. Then M lines follow, each gives the information of a person in the same format as the input, according to the ranking rules. If there is a tie of the total grade, they must be ranked with respect to their virtue grades in non-increasing order. If there is still a tie, then output in increasing order of their ID‘s.

　　Sample Input:
　　14 60 80
　　10000001 64 90
　　10000002 90 60
　　10000011 85 80
　　10000003 85 80
　　10000004 80 85
　　10000005 82 77
　　10000006 83 76
　　10000007 90 78
　　10000008 75 79
　　10000009 59 90
　　10000010 88 45
　　10000012 80 100
　　10000013 90 99
　　10000014 66 60

　　Sample Output:
　　12
　　10000013 90 99
　　10000012 80 100
　　10000003 85 80
　　10000011 85 80
　　10000004 80 85
　　10000007 90 78
　　10000006 83 76
　　10000005 82 77
　　10000002 90 60
　　10000014 66 60
　　10000008 75 79
　　10000001 64 90

参考代码：

 1 /****************************************************
 2 PAT B1015/A1062 Talent and Virtue
 3 ****************************************************/
 4 #include <iostream>
 5 #include <algorithm>
 6 #include <vector>
 7
 8 using namespace std;
 9
10 struct PeopleInfo {
11     string id;
12     int virtueGrade = 0;
13     int talentGRade = 0;
14     int totalGrade = 0;    //德智总分
15     int level = 0;         //等级：圣人--1;君子--2;愚人--3;小人--4
16 };
17
18 //自定义比较函数
19 bool myCmp(PeopleInfo A, PeopleInfo B) {
20     if (A.level == B.level) {
21         if (A.totalGrade == B.totalGrade) {
22             if (A.virtueGrade == B.virtueGrade) {
23                 return A.id < B.id;
24             }
25             else return A.virtueGrade > B.virtueGrade;
26         }
27         else return A.totalGrade > B.totalGrade;
28     }
29     else return A.level < B.level;
30 }
31
32 int main() {
33     vector<PeopleInfo> RankList;
34     int peopleCnt = 0, lowerLine = 0, hightLine = 0;
35
36     cin >> peopleCnt >> lowerLine >> hightLine;
37
38     //获取参与排名的人员信息
39     PeopleInfo temp;
40     for (int i = 0; i < peopleCnt; ++i) {
41         cin >> temp.id >> temp.virtueGrade >> temp.talentGRade;
42
43         temp.totalGrade = temp.virtueGrade + temp.talentGRade;
44
45         //计算等级
46         if (temp.virtueGrade < lowerLine || temp.talentGRade < lowerLine) continue;
47         else if (temp.virtueGrade >= hightLine && temp.talentGRade >= hightLine) temp.level = 1;
48         else if (temp.virtueGrade >= hightLine && temp.talentGRade < hightLine) temp.level = 2;
49         else if (temp.virtueGrade >= temp.talentGRade) temp.level = 3;
50         else temp.level = 4;
51
52         //达到排名标准的进入待排名容器
53         RankList.push_back(temp);
54     }
55
56     //对合格人员进行排名
57     sort(RankList.begin(), RankList.end(), myCmp);
58
59     //输出排名结果
60     cout << RankList.size() << endl;
61     for (int i = 0; i < RankList.size(); ++i) {
62         cout << RankList[i].id << ‘ ‘ << RankList[i].virtueGrade << ‘ ‘ << RankList[i].talentGRade;
63         if (1 != RankList.size() - 1) cout << endl;
64     }
65
66     return 0;
67 }

注意事项：

　　1：使用scanf和printf代替cin、cout会进一步减少程序的运行时间。

原文地址：https://www.cnblogs.com/mrdragon/p/11421190.html