标签：PAT int de cai else stu flag Virtue A1062
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 (≤105), 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
题意:
第一行给出三个整数：考生人数N，最低录取线L，优先录取线H
德分和才分均不低于L的才有录取资格
德才分均不低于H的为I类，最优先录取
德分不低于H但才分低于H的为II类，次优先录取
德分和才分均低于H但德分不低于才分的为III类，更次录取
其余考生为IV类，最次优先录取
思路：
1、排序的逻辑
①按类别从低到高排序
②类别相同，按总分从高到低排序
③总分相同，按德分从高到低排序
④德分相同，按考号从低到高排序
2、所需要的信息及信息的存储
①对于单个学生student，需要对应的准考证号id，德分de，才分cai，总分sum，类别flag——结构体Student，并创建一个足够大的结构体数组
②此外还需要总考生数n，最低分数线l，优先录取分数线h，录取考生数num——int型变量
3、排序逻辑的实现
1 bool cmp(Student a, Student b) {
2 if (a.flag != b.flag) {
3 return a.flag < b.flag;
4 } else if (a.sum != b.sum) {
5 return a.sum > b.sum;
6 } else if (a.de != b.de) {
7 return a.de > b.de;
8 } else {
9 return strcmp(a.id, b.id) < 0;
10 }
11 }
4、main() 雏形
int main() {
int n, l, h, num = 0;
scanf("%d%d%d", &n, &l, &h);
for (int i = 0; i < n; i++) {
scanf("%s%d%d", stu[i].id, &stu[i].de, &stu[i].cai);
stu[i].sum = stu[i].de + stu[i].cai;
if (stu[i].de < l || stu[i].cai < l) {
stu[i].flag = 5;
} else {
num++;
if (stu[i].de >= h && stu[i].cai >= h) {
stu[i].flag = 1;
} else if (stu[i].de >= h && stu[i].cai < h) {
stu[i].flag = 2;
} else if (stu[i] < h && stu[i].cai < h && stu[i].de >= stu[i].cai) {
stu[i].flag = 3;
} else {
stu[i].flag = 4;
}
}
}
sort(stu, stu + n, cmp);
}
5、优化
①逻辑优化
| 德分 < L | L <= 德分 < H | 德分 >= H | |
| 才分 < L | 5 | 5 | 5 |
| L <= 才分 < H | 5 | 3 | 2 |
| 才分 >= H | 5 | 4 | 1 |
所以，if-else逻辑应简化为
1 if (stu[i].de < l || stu[i].cai < l) {
2 stu[i].flag = 5;
3 } else if (stu[i].de >= h && stu[i].cai >= h) {
4 stu[i].flag = 1;
5 } else if (stu[i].de >= h && stu[i].cai < h) {
6 stu[i].flag = 2;
7 } else if (stu[i].de >= stu[i].cai) {
8 stu[i].flag = 3;
9 } else {
10 stu[i].flag = 4;
11 }
②num的替换
若在①中后四个else if中，每个都写上num++，会显得过于繁杂
因此将num初值设为参与考试的人数n，每次出现不合格的人时num--
1 int num = n;
2
3 if (stu[i].de < l || stu[i].cai < l) {
4 stu[i].flag = 5;
5 num--;
6 }
6、题解
1 #include<cstdio>
2 #include<cstring>
3 #include<algorithm>
4 using namespace std;
5 struct Student {
6 char id[10];
7 int de;
8 int cai;
9 int sum;
10 int flag;
11 }stu[100010];
12 bool cmp(Student a, Student b) {
13 if (a.flag != b.flag) {
14 return a.flag < b.flag;
15 } else if (a.sum != b.sum) {
16 return a.sum > b.sum;
17 } else if (a.de != b.de) {
18 return a.de > b.de;
19 } else {
20 return strcmp(a.id, b.id) < 0;
21 }
22 }
23 int main() {
24 int n, l, h;
25 scanf("%d%d%d", &n, &l, &h);
26 int num = n;
27 for (int i = 0; i < n; i++) {
28 scanf("%s%d%d", stu[i].id, &stu[i].de, &stu[i].cai);
29 stu[i].sum = stu[i].de + stu[i].cai;
30 if (stu[i].de < l || stu[i].cai < l) {
31 stu[i].flag = 5;
32 num--;
33 } else if (stu[i].de >= h && stu[i].cai >= h) {
34 stu[i].flag = 1;
35 } else if (stu[i].de >= h && stu[i].cai < h) {
36 stu[i].flag = 2;
37 } else if (stu[i].de >= stu[i].cai) {
38 stu[i].flag = 3;
39 } else {
40 stu[i].flag = 4;
41 }
42 }
43 sort(stu, stu + n, cmp);
44 printf("%d\n", num);
45 for (int i = 0; i < num; i++) {
46 printf("%s %d %d\n", stu[i].id, stu[i].de, stu[i].cai);
47 }
48 }
*7、未解决的问题
题目中准考证号给出8位
但实际使用char[8]存储再输出时会出错？？？
标签：PAT,int,de,cai,else,stu,flag,Virtue,A1062 来源： https://www.cnblogs.com/Bananice/p/12234582.html
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