【codeforces#25B】 Phone numbers

最新推荐文章于 2023-03-21 17:34:45 发布

Tiramicrystal

最新推荐文章于 2023-03-21 17:34:45 发布

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本文介绍了一种将电话号码按两位或三位分组的简单算法，以便于记忆。该算法适用于长度为2到100位的电话号码，并提供了一个C++实现示例。

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;
}