pat 1083. List Grades (25)

1083. List Grades (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

解：结构体存储+按成绩重新排序+控制输出

代码：

#include<iostream>
#include<cstdlib>
#include<cstdio>
using namespace std;
struct stu
{
    string name;
    string id;
    int grade;
};
int main()
{
    int n;
    while(scanf("%d",&n)==1)
    {
        stu *s=new stu[n+1];
        for(int i=0;i<n;i++)
        {
            cin>>s[i].name>>s[i].id>>s[i].grade;
        }
        int low,upp;
        scanf("%d%d",&low,&upp);
        for(int i=0;i<n;i++)
        {
            int maxi=i;
            for(int j=i+1;j<n;j++)
            {
                if(s[j].grade>s[maxi].grade)
                    maxi=j;
            }
            stu temp=s[maxi];
            s[maxi]=s[i];
            s[i]=temp;
        }
        if(s[0].grade<low||s[n-1].grade>upp)
        {
              cout<<"NONE"<<endl;
              break;
        }
        else
        {
            for(int i=0;i<n;i++)
            {
                if(s[i].grade>=low&&s[i].grade<=upp)
                {
                    cout<<s[i].name<<' '<<s[i].id<<endl;
                }
             }
        }
    }
}

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