List Grades (25)

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本篇博客介绍了一道编程题的解决方法，题目要求对学生的成绩按非递增顺序排序，并输出成绩位于给定区间内的学生记录。通过定义结构体存储学生信息并使用自定义比较函数进行排序。

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List Grades (25)

时间限制 1000 ms 内存限制 65536 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)
题目描述
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.

输入描述:
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.

输出描述:
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.

输入例子:
4
Tom CS000001 59
Joe Math990112 89
Mike CS991301 100
Mary EE990830 95
60 100

输出例子:
Mike CS991301
Mary EE990830
Joe Math990112

题解：
建立结构体，然后选择结构体进行排序就ok了。下面是AC代码

#include <bits/stdc++.h>
#include <stdlib.h>
using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define x first
#define y second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define SZ(x) x.size()
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define pi acos(-1)
#define mod 1000000007
#define inf 1000000007
#define ll long long
#define DBG(x) cerr<<(#x)<<"="<<x<<"\n";
#define N 200010
template <class U,class T> void Max(U &x, T y){if(x<y)x=y;}
template <class U,class T> void Min(U &x, T y){if(x>y)x=y;}
template <class T> void add(int &a,T b){a=(a+b)%mod;}
struct Stu{
    char name[10];
    char id[10];
    int grade;
}stu[N];
bool cmp(Stu a,Stu b)
{
    return a.grade>b.grade;
}
int main(){

   int n,g1,g2;
   bool f1=false;
   cin>>n;
   rep(i,0,n)
   {
      scanf("%s %s %d",stu[i].name,stu[i].id,&stu[i].grade);
   }
   scanf("%d%d",&g1,&g2);

   sort(stu,stu+n,cmp);

   rep(i,0,n)
   {
       if(stu[i].grade<=g2&&stu[i].grade>=g1)
       {
           f1=true;
           cout<<stu[i].name<<" "<<stu[i].id<<endl;
       }
   }
   if(f1==false)
   {

       cout<<"NONE"<<endl;
   }
   return 0;

}