【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

題意：給一串長度爲n的數字串，在兩個或三個數字之間加-，便於記憶。

解題思路：可以將數字串都3個3個分，遇到餘數不同的分情況討論。

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;
}