PAT甲级练习1083. List Grades (25)

最新推荐文章于 2021-12-04 10:16:58 发布

原创最新推荐文章于 2021-12-04 10:16:58 发布 · 52 阅读

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本文介绍了一个简单的程序设计问题，即按非递增顺序对学生记录进行排序，并筛选出成绩位于指定区间内的学生记录。

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

没什么难度

#include <iostream>  
#include <cstdio>  
#include <algorithm>  
#include <vector>  
#include <map>  
#include <set>  
#include <stack>  
#include <queue>  
#include <string>  
#include <string.h> 
using namespace std;

const int inf = 99999999;
const int MAX = 1e5+10;

struct student{
  char name[20];
  char id[20];
  int grade;
}s[MAX];

bool cmp(student a, student b){
  return a.grade>b.grade;
}

int main() {
  int n, g1, g2, cnt=0;
  scanf("%d", &n);

  for(int i=0; i<n; i++){
    scanf("%s %s %d", s[i].name, s[i].id, &s[i].grade);
  }
  sort(s, s+n, cmp);
  scanf("%d %d", &g1, &g2);
  for(int i=0; i<n; i++){
    if(s[i].grade>=g1 && s[i].grade<=g2){
      printf("%s %s\n", s[i].name, s[i].id);
      cnt++;
    }
  }
  if(cnt==0) printf("NONE\n");
  scanf("%d",&n);
    return 0;
}