A1083 List Grades (25 分)
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
开始技术总结:
第一点就是初始化数组或是结构体的时候,大小要用常量,不能够使用输入量初始化
还有就是中文的括号注意一点
参考代码:
#define _CRT_SECURE_NO_WARNINGS
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 50;
struct Student {
char ID[11];
char name[11];
int grade;
}student[MAXN];
bool cmp(Student a, Student b) {
if (a.grade != b.grade) return a.grade > b.grade;
}
int main(){
int grade1, grade2;
int N;
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s%s%d", student[i].name, student[i].ID, &student[i].grade);
}
scanf("%d%d", &grade1, &grade2);
sort(student, student + N, cmp);
int flag = 0;//用来标记是否又符合条件的同学,0表示没有
for (int j = 0; j < N; j++) {
if (student[j].grade <= grade2 && student[j].grade >= grade1) {
printf("%s %s\n", student[j].name, student[j].ID);
flag = 1;
}
}
if (flag == 0) {
printf("NONE");
}
system("pause");
return 0;
}
A1083 List Grades (25 分)的更多相关文章
- A1083 List Grades (25)(25 分)
A1083 List Grades (25)(25 分) Given a list of N student records with name, ID and grade. You are supp ...
- 【PAT甲级】1083 List Grades (25 分)
题意: 输入一个正整数N(<=101),接着输入N个学生的姓名,id和成绩.接着输入两个正整数X,Y(0<=X,Y<=100),逆序输出成绩在x,y之间的学生的姓名和id. tric ...
- PAT 甲级 1083 List Grades (25 分)
1083 List Grades (25 分) Given a list of N student records with name, ID and grade. You are supposed ...
- 1012 The Best Rank (25 分)
1012 The Best Rank (25 分) To evaluate the performance of our first year CS majored students, we cons ...
- 1062 Talent and Virtue (25 分)
1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a history ...
- 1036 Boys vs Girls (25 分)
1036 Boys vs Girls (25 分) This time you are asked to tell the difference between the lowest grade of ...
- A1012 The Best Rank (25)(25 分)
A1012 The Best Rank (25)(25 分) To evaluate the performance of our first year CS majored students, we ...
- PATA1012The Best Rank(25分)
To evaluate the performance of our first year CS majored students, we consider their grades of three ...
- PAT 甲级 1062 Talent and Virtue (25 分)(简单,结构体排序)
1062 Talent and Virtue (25 分) About 900 years ago, a Chinese philosopher Sima Guang wrote a histor ...
随机推荐
- Spark学习笔记01-基础
目录 简介 特性 Spark运行模式 Mac本地安装 本文基于 Spark 2.4.1 进行演示,相关代码可以在我的Github上看到. 简介 Spark是一个分布式集群计算系统,类似Hadoop提供 ...
- spring boot 不占用端口方式启动
随着微服务架构的流行,想要启动一个微服务架构项目就要开启好多端口,有时候一台机器上部署的项目多的时候,端口资源就比较紧张了,其实有的微服务组件仅仅只是提供RPC服务,可以不用占用web启动的端口,此时 ...
- set -x 与 set +x
set -x 与 set +x 在liunx脚本中可用set -x就可有详细的日志输出.免的老是要echo了 下面的网上搜来的用法. 用于脚本调试.set是把它下面的命令打印到屏幕set -x 是开启 ...
- Linux和windows下修改tomcat内存
原文地址:https://www.cnblogs.com/wdpnodecodes/p/8036333.html 由于服务器上放的tomcat太多,造成内存溢出. 常见的内存溢出有以下两种: java ...
- LeetCode 142:环形链表 II Linked List Cycle II
给定一个链表,返回链表开始入环的第一个节点. 如果链表无环,则返回 null. 为了表示给定链表中的环,我们使用整数 pos 来表示链表尾连接到链表中的位置(索引从 0 开始). 如果 pos 是 - ...
- playtime-浙大羽协裁判部训练方案[随机事件序列的应用]
首先随机一列人名 然后按比例随机一列事件项. 然后将不确定项的人名更正为“某人”[比如发球违例,,,你怎么知道谁在发球] 最后定义一个初始化. 初始化呢,就是挑边. 球权还是场权? 发球还是接发? 谁 ...
- 【MySQL】Mariadb主从复制
Mariadb主从复制 环境配置: Master : 172.30.200.200 Slave : 172.30.200.204 架构图,如下: Master的配置: 1.binlog配置 [root ...
- 怎样用 C# 快速比较 2 个文件是否是相同的文件?
方案1: 直接贴代码了: using System; using System.Collections.Generic; using System.IO; using System.Linq; usi ...
- MySQL for OPS 02:SQL 基础
写在前面的话 上一节主要谈谈 MySQL 是怎么安装的以及最简单的初始化我们应该做哪些配置.其中也用到了一些简单的用户操作 SQL,所以这一节主要学习常用的 SQL 使用. SQL 介绍 在了解 SQ ...
- conda opencv cv2.imshow无法使用
error: -------src-dir-------/opencv-2.4.10/modules/highgui/src/window.cpp:501: error: (-2) The funct ...