B1028 census

1028 The census (20 branch )

A census was conducted in a town , Got all the residents' birthdays . Now please write a program , Find out the oldest and youngest people in town .

Make sure that every date you enter is legal , But it's not necessarily reasonable —— Suppose there's no more than 200 Old man , And today is 2014 year 9 month 6 Japan , So over 200 Birthday and unborn birthday are unreasonable , It should be filtered out .

Input format ：
The input gives a positive integer on the first line N, The value is (0,10 Of 5 Power ]; And then N That's ok , Each line gives 1 Personal name （ By no more than 5 A string of English letters ）、 And press yyyy/mm/dd（ That is the year / month / Japan ） Birthday given in format . The title ensures that the oldest and youngest people are not tied .

Output format ：

Output the number of valid birthdays in order in one line 、 The names of the oldest and youngest people , Separated by spaces .

sample input

5
John 2001/05/12
Tom 1814/09/06
Ann 2121/01/30
James 1814/09/05
Steve 1967/11/20

sample output

3 Tom John

Topic analysis ：

1. After creating the structure, it is very simple

The code is as follows ：

#include<bits/stdc++.h>
using namespace std;
struct people{
    string name;
    int old;
}p[100100];
int main(){
    int n,count=0;
    string s;
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>p[i].name>>s;
        p[i].old=(s[0]-'0')*10000000+(s[1]-'0')*1000000+(s[2]-'0')*100000+(s[3]-'0')*10000+(s[5]-'0')*1000+(s[6]-'0')*100+(s[8]-'0')*10+(s[9]-'0');
        if(p[i].old<=20140906&&20140906-p[i].old<=2000000){
            count++;
        }
    }
    if(count==0){
        cout<<"0"<<endl;
        return 0;
    }
    int max=0,min=100000000;
    for(int i=0;i<n;i++){
        if(p[i].old<=20140906&&20140906-p[i].old<=2000000){
            if(p[i].old<min){
                min=p[i].old;
            }
            if(p[i].old>max){
                max=p[i].old;
            }
        }
    }
    for(int i=0;i<n;i++){
        if(min==p[i].old){
            min=i;
        }
        if(max==p[i].old){
            max=i;
        }
    }
    cout<<count<<" "<<p[min].name<<" "<<p[max].name<<endl;
}