【codeforces#25B】 Phone numbers

B. Phone numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Phone number in Berland is a sequence of n digits. Often, to make it easier to memorize the number, it is divided into groups of two or three digits. For example, the phone number 1198733 is easier to remember as 11-987-33. Your task is to find for a given
phone number any of its divisions into groups of two or three digits.

Input

The first line contains integer n (2 ≤ n ≤ 100) — amount of digits in the phone number. The second line contains n digits — the phone number to divide into groups.

Output

Output any of divisions of the given phone number into groups of two or three digits. Separate groups by single character -. If the answer is not unique, output any.

Sample test(s)

Input

6

549871

Output

54-98-71

Input

7

1198733

Output

11-987-33

题意：给一串长度为n的数字串，在两个或三个数字之间加-，便于记忆。

解题思路：可以将数字串都3个3个分，遇到余数不同的分情况讨论。

code：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
using namespace std;
int main()
{
int n;
string ss;
cin>>n>>ss;
if(n%3==0)
{
printf("%c%c%c",ss[0],ss[1],ss[2]);
for(int i=3;i<n;i+=3){
printf("-%c%c%c",ss[i],ss[i+1],ss[i+2]);
}
printf("\n");
}
else if(n%3==1)
{
printf("%c%c-%c%c",ss[0],ss[1],ss[2],ss[3]);
for(int i=4;i<n;i+=3)
printf("-%c%c%c",ss[i],ss[i+1],ss[i+2]);
printf("\n");
}
else
{
printf("%c%c",ss[0],ss[1]);
for(int i=2;i<n;i+=3)
printf("-%c%c%c",ss[i],ss[i+1],ss[i+2]);
printf("\n");
}
return 0;
}