sgu 127 Telephone directory

127. Telephone directory

time limit per test: 0.5 sec.

memory limit per test: 4096 KB
CIA has decided to create a special telephone directory for its agents. The first 2 pages of the directory contain the name of the directory and instructions for agents, telephone number records begin on the third
page. Each record takes exactly one line and consists of 2 parts: the phone number and the location of the phone. The phone number is 4 digits long. Phone numbers cannot start with digits 0 and 8. Each page of the telephone directory can contain not more then
K lines. Phone numbers should be sorted in increasing order. For the first phone number with a new first digit, the corresponding record should be on a new page of the phone directory. You are to write a program, that calculates the minimal number P pages
in the directory. For this purpose, CIA gives you the list of numbers containing N records, but since the information is confidential, without the phones locations.

Input
The first line contains a natural number K (0 < K < 255) - the maximum number of lines that one page can contain. The second line contains a natural N (0 < N < 8000) - number of phone numbers supplied. Each of
following N lines contains a number consisting of 4 digits - phone numbers in any order, and it is known, that numbers in this list cannot repeat.
Output
First line should contain a natural number P - the number of pages in the telephone directory.
Sample Input

5
10
1234
5678
1345
1456
1678
1111
5555
6789
6666
5000

Sample Output

5

很裸的暴力模拟一下 ，其实真的是水题。
代码：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<set>
#include<algorithm>
#include<vector>
#include<cstdlib>

#define inf 0xfffffff
#define CLR(a,b) memset((a),(b),sizeof((a)))

using namespace std;
int const nMax = 4000;
typedef int LL;
typedef pair<LL,LL> pij;

int a[10];
char s[20];
int k,n;

int main(){
scanf("%d",&k);
scanf("%d",&n);
for(int i=0;i<n;i++){
scanf("%s",s);
a[s[0]-'0']++;
}
int ans=2;
for(int i=1;i<10;i++){
if(i==8)continue;
ans+=a[i]/k;
if(a[i]%k)ans++;
}
printf("%d\n",ans);
return 0;
}