List Grades(PAT)

1.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. 给出N个学生的名字，ID和年级的记录列表。您应该按不递增的顺序对成绩进行排序，并输出成绩在给定间隔内的学生记录。

2.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. 每个输入文件包含一个测试用例。每个案例的格式如下： N name[1] ID[1] grade[1] name[2] ID[2] grade[2] … … name[N] ID[N] grade[N] grade1 grade2 其中name[i]和ID[i]是不超过10个字符的字符串，没有空格，grade[i]是[0,100]中的整数，grade1 grade2是级别间隔的边界。保证所有年级都是不同的。

3.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. 对于每个测试用例，您应该输出学生记录，其中的成绩在给定的间隔内[grade1, grade2]，并且是按不递增的顺序排列的。每个学生的记录与学生的姓名和ID占据一行，用一个空格隔开。如果在这段时间内没有学生的成绩，则输出“NONE”。

4.输入例子:

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

5.输出例子:

Mike CS991301 Mary EE990830 Joe Math990112

6.源代码：

#include<stdio.h> typedef struct { char name[10]; char id[10]; int grade; }student; void qsort(student stu[],int L,int R)//快速排序 { if(L<R) { int i=L; int j=R; student key=stu[L]; while(i<j) { while(i<j&&stu[j].grade<=key.grade) j--; if(i<j) stu[i++]=stu[j]; while(i<j&&stu[i].grade>key.grade) i++; if(i<j) stu[j--]=stu[i]; } stu[i]=key; qsort(stu,L,i-1); qsort(stu,i+1,R); } } int main() { int i,count=0,N; //输入 scanf("%d",&N); student stu[101]; for(i=0;i<N;i++) scanf("%s %s %d",stu[i].name,stu[i].id,&stu[i].grade); int grade1,grade2; scanf("%d %d",&grade1,&grade2); //排序 qsort(stu,0,N-1); //输出 for(i=0;i<N;i++) if(stu[i].grade>=grade1&&stu[i].grade<=grade2) { count++; printf("%s %s\n",stu[i].name,stu[i].id); } //对符合条件的学生计数，如果没有就输出NONE if(count==0) printf("NONE\n"); return 0; }