Address book sorting

Input n A friend's information , Including name, 、 Birthday 、 Phone number , This question requires the preparation of procedures , Output the address book in order of age . The title ensures that everyone's birthday is different .

Input format :

Enter the first line to give a positive integer n（<10）. And then n That's ok , Each line follows “ full name Birthday Phone number ” Give a friend's information in the format of , among “ full name ” It's no longer than 10 A string of English letters ,“ Birthday ” yes yyyymmdd Format date ,“ Phone number ” No more than 17 Digit numbers and +、- Composed string .

Output format :

Output your friends' information according to their age , The format is the same as that of output .

sample input :

3
zhang 19850403 13912345678
wang 19821020 [phone]
qian 19840619 13609876543

sample output :

wang 19821020 [phone]
qian 19840619 13609876543
zhang 19850403 13912345678

Code ： 

#include <stdio.h>
#include <stdlib.h>
struct friend
{
    char name[10];
    int brithday;
    char phone[17];
};
int main()
{
    int n,i,j;
    int min;
    struct friend a[11];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("\n");
        scanf("%s %d %s",&a[i].name,&a[i].brithday,&a[i].phone);
    }
    for(i=0;i<n;i++)
    {
        for(j=0;j<n-1-i;j++)
        {
            if(a[j].brithday>a[j+1].brithday)
            {
                a[10]=a[j];
                a[j]=a[j+1];
                a[j+1]=a[10];
            }
        }
    }
    for(i=0;i<n;i++)
    {
        printf("%s %d %s\n",a[i].name,a[i].brithday,a[i].phone);
    }
    return 0;
}