组队选拔赛01 ---- trener

Problem Description

Mirko has been moving up in the world of basketball, starting as a mere spectator, mastering snack salesmanship, finally reach the coveted position of the national team coach. He is now facing a difficult

task: selecting the five primary players for the upcoming match against Tajikistan.Since Mirko is incredibly lazy, he doesn’t bother remembering players’ names, let alone their actual skills. That’s why he has settled on selecting five players who share the same first letter of their surnames, so that he can remember them more easily. If there are no five players sharing the first letter of their surnames, Mirko will simply forfeit the game!

In order to obtain insight into possibilities for his team, Mirko wants to know all the different letters that his primary team’s surnames may begin with.

Input

There are multiple test cases. Please process till EOF.

The first line of input contains the positive integer N (1 ≤ N ≤ 150), the number of players that Mirko has available.

Each of the following N lines contains one word (at most 30 characters long, consisting only of lowercase English letters), a surname of one of the players.

Output

If there are no five players that Mirko can select matching his criteria, output a single line containing the word “PREDAJA” (without quotes). Otherwise, output all possible first letters of representation player surnames, sorted lexicographically, in a single line with no spaces.

Sample Input

18

babic

keksic

boric

bukic

sarmic

balic

kruzic

hrenovkic

beslic

boksic

krafnic

pecivic

klavirkovic

kukumaric

sunkic

kolacic

kovacic

prijestolonasljednikovic

6

michael

jordan

lebron

james

kobe

bryant

Sample Output

bk

PREDAJA

解题思路

题意就是 找出名字首字母相同且人数大等于5的 字母.

用map的话可以映射字母和当前的人数,再判断是否大等于5即可.

参考代码

#include <cstdio>
#include <map>
using namespace std;
int main()
{
int n;
char name[35];
while (~scanf("%d", &n)){
map<char,int> m;
while (n--){
scanf("%s", name);
m[name[0]]++;
}
bool find = false;
for (int i = 'a';i <= 'z';i++)
if (m[i] >= 5){
printf("%c",i);
find = true;
}
if (!find)  printf("PREDAJA");
printf("\n");
}
return 0;
}