1028 Census (20 minutes)
A town census, was the birthday of all residents. Now you write a program to find the town's oldest and most young people.
Make sure that each date entered here are legitimate, but not necessarily reasonable - assuming there is no known town more than 200 years old, and today is September 6, 2014, the more than 200-year-old's birthday and not born birthday is unreasonable and should be filtered out.
Input formats:
In the first line of the input is given a positive integer N, the values (0,10 5]; followed by N rows, each individual is given a name (a string of not more than 5 letters of the English), and by yyyy/mm/ddbirth (ie year / month / day) format given in the title to ensure the oldest and the youngest person not tied.
Output formats:
The number of valid order of the output birthday in a row, young people and most of the oldest names, separated by spaces between them.
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
Ideas: Note special sentenced reasonable number is zero.
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
struct Node{
char name[109];
int yy,mm,dd;
}a[100009];
int cmp( struct Node a ,struct Node b ){
if ( a.yy != b.yy ){
return a.yy < b.yy;
}
if( a.mm != b.mm ){
return a.mm <b.mm;
}
return a.dd < b.dd;
}
bool check( int yy ,int mm ,int dd ){
if( yy< 1814 )
return false;
else if( yy == 1814 ){
if( mm < 9 ){
return false;
}
else if( mm == 9 ){
if( dd <6 ){
return false;
}
}
}
else if( yy > 2014 )
return false;
else if( yy ==2014 ){
if( mm >9 ){
return false;
}
else if( mm == 9 ){
if( dd >6 ){
return false;
}
}
}
return true;
}
int main(void){
int n;
while( scanf("%d",&n)!=EOF){
int ans = 0,yy,mm,dd;
char name[109];
for( int i=0;i<n;i++){
scanf("%s",name);
scanf("%d",&yy);
getchar();
scanf("%d",&mm);
getchar();
scanf("%d",&dd);
if( check(yy,mm,dd) ){
strcpy(a[ans].name,name);
a[ans].yy = yy;
a[ans].mm = mm;
a[ans].dd = dd;
ans++;
}
}
sort(a,a+ans,cmp);
if( ans ==0 ){
printf("0\n");
}
else printf("%d %s %s",ans,a[0].name,a[ans-1].name);
}
return 0;
}