某城镇进行人口普查,得到了全体居民的生日。现请你写个程序,找出镇上最年长和最年轻的人。
这里确保每个输入的日期都是合法的,但不一定是合理的——假设已知镇上没有超过 200 岁的老人,而今天是 2014 年 9 月 6 日,所以超过 200 岁的生日和未出生的生日都是不合理的,应该被过滤掉。
输入格式:
输入在第一行给出正整数 N,取值在(0,10];随后 N 行,每行给出 1 个人的姓名(由不超过 5 个英文字母组成的字符串)、以及按 yyyy/mm/dd(即年/月/日)格式给出的生日。题目保证最年长和最年轻的人没有并列。
输出格式:
在一行中顺序输出有效生日的个数、最年长人和最年轻人的姓名,其间以空格分隔。
输入样例:
5John 2001/05/12Tom 1814/09/06Ann 2121/01/30James 1814/09/05Steve 1967/11/20
输出样例:
3 Tom John
思路
这题面向对象思想浓厚,第一次用了Java来写
第4个测试点没过。。。。
第二次回顾用C++写,代码AC。
代码
import java.util.Scanner;class Person {public String name;public int year;public int month;public int day;public Person() {}public Person(String name, int year, int month, int day) {this.name = name;this.year = year;this.month = month;this.day = day;}public boolean Bigger(Person left) {if(this.year != left.year)return this.year >= left.year;else if(this.month != left.month)return this.month >= left.month;elsereturn this.day >= left.day;}public boolean Smaller(Person right) {if(this.year != right.year)return this.year <= right.year;else if(this.month != right.month)return this.month <= right.month;elsereturn this.day <= right.day;}}public class Main {public static void main(String[] args) {/** Iinitialize the data */Person OLD = new Person("OLD", 2014, 9, 6);Person YOUNG = new Person("YOUNG", 1814, 9, 6);Person left = new Person("Left", 1814, 9, 6);Person right = new Person("Right", 2014, 9, 6);/** Input the number */int number, counter = -1;Scanner input = new Scanner(System.in);number = input.nextInt();/** Input the person information */Person[] person = new Person[10];for(int i = 0; i < number; i++) {person[i] = new Person();}String tempDate = new String();for(int i = 0; i < number; i++) {person[i].name = input.next();tempDate = input.next();formatInput(person[i], tempDate);// Fliter the invalid dataif(person[i].Bigger(left) && person[i].Smaller(right)) {counter++;// Update OLD and YOUNGif(person[i].Smaller(OLD)) {OLD = person[i];}if(person[i].Bigger(YOUNG)) {YOUNG = person[i];}}}/** Display the result */if(counter == -1) {System.out.println("0");}elseSystem.out.printf("%d %s %s", counter + 1, OLD.name, YOUNG.name);}public static void formatInput(Person temp, String tempDate) {// Cut the bitsString date[] = tempDate.split("/");String year = date[0];String month = date[1];String day = date[2];// Convert String into integerint iYear = Integer.parseInt(year);int iMonth = Integer.parseInt(month);int iDay = Integer.parseInt(day);// Assign the integers into objecttemp.year = iYear;temp.month = iMonth;temp.day = iDay;}}
AC代码
#include <iostream>using namespace std;typedef struct Person {char name[10];int year;int month;int day;}Person;Person OLD, YOUNG, leftEdge, rightEdge;void init();bool LessEqu(Person& a, Person& b);bool GreaterEqu(Person& a, Person& b);int main() {/* Initialize edge */init();/* Read data */int N, counter = 0;cin >> N;for(int i = 0; i < N; i++) {Person tmp;scanf("%s %d/%d/%d", &tmp.name, &tmp.year,&tmp.month, &tmp.day);/* Check tmp is valid or not? */if( GreaterEqu(tmp, leftEdge) && LessEqu(tmp, rightEdge) ) {counter++;if( LessEqu(tmp, OLD) )OLD = tmp;if( GreaterEqu(tmp, YOUNG) )YOUNG = tmp;}}if(counter == 0)printf("0\n");elseprintf("%d %s %s\n", counter, OLD.name, YOUNG.name);return 0;}void init() {YOUNG.year = leftEdge.year = 1814;OLD.year = rightEdge.year = 2014;YOUNG.month = OLD.month = 9;leftEdge.month = rightEdge.month = 9;YOUNG.day = OLD.day = 6;leftEdge.day = rightEdge.day = 6;}bool LessEqu(Person& a, Person& b) {if( a.year != b.year )return a.year <= b.year;else if( a.month != b.month )return a.month <= b.month;elsereturn a.day <= b.day;}bool GreaterEqu(Person& a, Person& b) {if( a.year != b.year )return a.year >= b.year;else if( a.month != b.month )return a.month >= b.month;elsereturn a.day >= b.day;}
