Algorithm notes on combat training aircraft Beginners Guide Chapter 4 (2) - 4.1 sorting algorithm preliminary study notes

PAT B1015 moral theory (25 points) / A1062

Song Dynasty historian Sima Guang, there is a famous "moral theory" in the "Mirror": "The actual occurrence of that of the saints to do the whole virtuous, virtuous and death that of a fool, Ruby was that of a gentleman, just Shengde that of the villain. people who take the surgery, Gou not saints, and with the gentleman, was its villain, was not as fools. "

Now gives moral and scores a number of candidates, please give admission ranked according to Sima Guang theory.

Input formats:

Input of the first line gives three positive integers, respectively:

Then

Output formats:

Firstly, a first output line to reach the minimum score of the number of candidates to

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

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct student{
    char id[10];
    int de,cai,sum;
    int flag; //分类 
}stu[100010];
bool cmp(student a,student b){
    if(a.flag != b.flag ) 
        return a.flag < b.flag;
    else if(a.sum != b.sum) 
        return a.sum > b.sum;
    else if(a.de != b.de) 
        return a.de > b.de;
    else
        return strcmp(a.id,b.id) < 0;
}
int main(){
    int N,L,H;
    scanf("%d%d%d",&N,&L,&H);
    int m = N;
    for(int i=0;i<N;i++){
        scanf("%s %d %d",stu[i].id,&stu[i].de,&stu[i].cai);
        this [the] .sum = this [i] + .de this [the] .cai;
        if (this [the] .de <L || this [the] .cai < L) {
            stu[i].flag = 5;
            m--;
        }else 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 (this [the] .de> = this [the] .cai) {
            stu[i].flag = 3;
        }else{
            stu[i].flag = 4;
        }
    }
    sort(stu,stu+N,cmp);
    printf("%d\n",m);
    for(int i = 0;i<m;i++){
        printf("%s %d %d\n",stu[i].id,stu[i].de,stu[i].cai);
    }

    return 0;
}

PAT A1012 The Best Rank (25分)

To evaluate the performance of our first year CS majored students, we consider their grades of three courses only: C - C Programming Language, M - Mathematics (Calculus or Linear Algrbra), and E - English. At the mean time, we encourage students by emphasizing on their best ranks -- that is, among the four ranks with respect to the three courses and the average grade, we print the best rank for each student.

For example, The grades of C, M, E and A - Average of 4 students are given as the following:

StudentID  C  M  E  A
310101     98 85 88 90 310102 70 95 88 84 310103 82 87 94 88 310104 91 91 91 91

Then the best ranks for all the students are No.1 since the 1st one has done the best in C Programming Language, while the 2nd one in Mathematics, the 3rd one in English, and the last one in average.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 2 numbers

Output Specification:

For each of the

The priorities of the ranking methods are ordered as A $> C > M > E . Hence if there are two or more ways for a student to obtain the same best rank, output the one with the highest priority.$

If a student is not on the grading list, simply output N/A.

Sample Input:

5 6
310101 98 85 88
310102 70 95 88
310103 82 87 94
310104 91 91 91
310105 85 90 90
310101
310102
310103
310104
310105
999999

Sample Output:

1 C
1 M
1 E
1 A
3 A
N/A

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct Student{
    int id;
    int score[4];
}stu[2010];
int now;
int Rank[1000000][4]={0};
char course[4]= {'A','C','M','E'};
bool cmp(Student a,Student b){
    return a.score[now] > b.score[now];
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d %d %d %d",&stu[i].id,&stu[i].score[1],&stu[i].score[2],&stu[i].score[3]);
        stu[i].score[0] = round((stu[i].score[1] + stu[i].score[2] + stu[i].score[3])/3.0);
    }
    for(now=0;now<4;now++){
        sort(stu,stu+n,cmp);
        Rank[stu[0].id][now] = 1;
        for(int i=1;i<n;i++){
            if(stu[i].score[now] == stu[i-1].score[now]){
                Rank[stu[i].id][now] = Rank[stu[i-1].id][now];
            }else{
                Rank[stu[i].id][now] = i + 1;
            }
        }
    }
    for(int i=0;i<m;i++){
        int query;
        scanf("%d",&query);
        if(Rank[query][0] == 0){
            printf("N/A\n");
        }else{
            int k=0;
            for(int j=0;j<4;j++){
                if(Rank[query][j] < Rank[query][k]){
                    k = j;
                }
            }
            printf("%d %c\n",Rank[query][k],course[k]);
        }
    }
    return 0;
}