PAT 乙級 1028 人口普查
阿新 • • 發佈:2019-02-04
直接利用字串的大小比較來比較日期的大小
#include <iostream>
using namespace std;
int main(){
int n;
cin >> n;
int count = 0;
string effeYoung = "2014/09/06", effeOld = "1814/09/06";
string old, young, oDate = effeYoung, yDate = effeOld;
string name, date;
for (int i=0; i<n; i++) {
cin >> name >> date ;
// 直接利用字串的大小比較來比較日期的大小
if (date <= effeYoung && date >= effeOld) {
count++;
if (date > yDate){
young = name;
yDate = date;
}
if (date < oDate) {
old = name;
oDate = date ;
}
}
}
cout << count;
if (count != 0)
cout << ' ' << old << ' ' << young << endl;
return 0;
}