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(0,105];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:
5 John 2001/05/12 Tom 1814/09/06 Ann 2121/01/30 James 1814/09/05 Steve 1967/11/20
输出样例:
3 Tom John
在提交时发生了一件神奇的事情,有时候提交部分正确,有时提交完全正确。//测试用例应该是会变的。
#include<iostream> #include<string> using namespace std; class birth { public: int year; int month; int day; bool isbirth() { return ((year < 2014 || (year == 2014 && month < 9) || (year == 2014 && month == 9 && day <= 6)) && (year > 1814 || (year == 1814 && month > 9) || (year == 1814 && month == 9 && day >= 6))); } bool operator>(birth b2) { return (year > b2.year || (year == b2.year && month > b2.month) || (year == b2.year && month == b2.month && day > b2.day)); } }; class person { public: birth birthday; string name; }; int main() { int N; person young; young.birthday.year = 2020; person old; int count=0; cin >> N; person tmp; for (int i = 0; i < N; i++) { cin >> tmp.name; scanf("%d/%d/%d", &tmp.birthday.year, &tmp.birthday.month, &tmp.birthday.day); if (tmp.birthday.isbirth()) { count++; if (tmp.birthday > old.birthday) old = tmp; if (young.birthday > tmp.birthday ) young= tmp; } } cout << count; if (count != 0) { cout << " " << young.name << " " << old.name; } return 0; }
想了一下,做了一点修改,当把第一个合法的人给young 和old,(因为这时只有一个人,他同时是最年轻和最老的)
#include<iostream> #include<string> using namespace std; class birth { public: int year; int month; int day; bool isbirth() { return ((year < 2014 || (year == 2014 && month < 9) || (year == 2014 && month == 9 && day <= 6)) && (year > 1814 || (year == 1814 && month > 9) || (year == 1814 && month == 9 && day >= 6))); } bool operator>(birth b2) { return (year > b2.year || (year == b2.year && month > b2.month) || (year == b2.year && month == b2.month && day > b2.day)); } }; class person { public: birth birthday; string name; }; int main() { int N; person young; person old; int count=0; cin >> N; person tmp; for (int i = 0; i < N; i++) { cin >> tmp.name; scanf("%d/%d/%d", &tmp.birthday.year, &tmp.birthday.month, &tmp.birthday.day); if (tmp.birthday.isbirth()) { if (count == 0) { old = tmp; young = tmp; } count++; if (tmp.birthday > old.birthday) old = tmp; if (young.birthday > tmp.birthday ) young= tmp; } } cout << count; if (count != 0) { cout << " " << young.name << " " << old.name; } return 0; }