1083. List Grades (25)

题目链接：http://www.patest.cn/contests/pat-a-practise/1083
题目：

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
 ID
 grade

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

分析：
结构体，排序，输出指定区间的元素

AC代码：

#include<stdio.h>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
struct Stu{
 string name;
 string id;
 int grade;
};
vector<Stu>V_Stu;
bool cmp(const Stu &A, const Stu &B){ //这个const一定要加上
 return A.grade > B.grade;
}
int main(){
 freopen("F://Temp/input.txt", "r", stdin);
 int n;
 cin >> n;
 V_Stu.clear();
 while (n--){
  Stu tmp_Stu;
  cin >> tmp_Stu.name >> tmp_Stu.id >> tmp_Stu.grade;
  V_Stu.push_back(tmp_Stu);
 }
 sort(V_Stu.begin(), V_Stu.end(), cmp);
 int start, end;
 int idx_s = -1, idx_t = -1;
 cin >> start >> end;
 for (int i = 0; i < V_Stu.size(); ++i){
  if (V_Stu[i].grade <= end){
   idx_s = i;
   break;
  }
 }
 for (int i = V_Stu.size() - 1; i >= 0; --i){
  if (V_Stu[i].grade >= start){
   idx_t = i;
   break;
  }
 }
 if (idx_s == -1 || idx_t == -1){
  puts("NONE");
  return 0;
 }
 for (int i = idx_s; i <= idx_t; i++){
  cout << V_Stu[i].name << " " << V_Stu[i].id << endl;
 }
 return 0;
}

截图：



——Apie陈小旭