PAT-A1083 List Grades 题目内容及题解

Given a list of N student records with name, ID and grade. You are supposed to sort the records with respect to the grade in non-increasing order, and output those student records of which the grades are in a given interval.

Input Specification:

Each input file contains one test case. Each case is given in the following format:

N
name[1] ID[1] grade[1]
name[2] ID[2] grade[2]
... ...
name[N] ID[N] grade[N]
grade1 grade2

where name[i] and ID[i] are strings of no more than 10 characters with no space, grade[i] is an integer in [0, 100], grade1 and grade2 are the boundaries of the grade's interval. It is guaranteed that all the grades are distinct.

Output Specification:

For each test case you should output the student records of which the grades are in the given interval [grade1, grade2] and are in non-increasing order. Each student record occupies a line with the student's name and ID, separated by one space. If there is no student's grade in that interval, output NONE instead.

Sample Input 1:

4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100

Sample Output 1:

Mike CS991301
Mary EE990830
Joe Math990112

Sample Input 2:

2
Jean AA980920 60
Ann CS01 80
90 95

Sample Output 2:

NONE

题目大意

题目给出一个包含若干学生记录的列表，其中包含姓名、ID和成绩。题目要求按照非递增的顺序对与成绩相关的记录进行排序，并输出成绩在给定间隔内的学生记录。

解题思路

1. 建立结构体保存数据；
2. 读入数据并储存；
3. 筛选其中符合要求的数据并标记、计数；
4. 按照题目要求排序；
5. 按照题目要求输出结果并返回零值。

代码

#include<cstdio>
#include<algorithm>
#include<vector> 
using namespace std;
struct Student{
    char name[12],id[12];
    int grade;
    int vivid;
};

bool cmp(Student a,Student b){
    if(a.vivid!=b.vivid){
        return a.vivid>b.vivid;
    }else{
        return a.grade>b.grade;
    }
}

int main(){
    int N,low,high;
    int i,cnt=0;
    Student temp;
    vector<Student> stu;
    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%s %s %d",&temp.name,&temp.id,&temp.grade);
        stu.push_back(temp);
    }
    scanf("%d%d",&low,&high);
    for(i=0;i<N;i++){
        if(stu[i].grade>=low&&stu[i].grade<=high){
            cnt++;
            stu[i].vivid=1;
        }
    }
    sort(stu.begin(),stu.end(),cmp);
    if(cnt==0){
        printf("NONE\n");
    }
    for(i=0;i<cnt;i++){
        printf("%s %s\n",stu[i].name,stu[i].id);
    }
    return 0;
} 

运行结果