B. Phone numbers
2014-05-21 15:24
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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
output
input
output
解题说明:模拟题,按照要求输出即可,可以每两位输出一次,最后输出三位即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
#include<string>
using namespace std;
int main()
{
int n,i;
string a;
cin >> n;
cin >> a;
for (i = 0; i<n; i++)
{
cout << a[i];
if (i % 2 && n - i - 1>1)
{
cout << '-';
}
}
cout << endl;
return 0;
}
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
解题说明:模拟题,按照要求输出即可,可以每两位输出一次,最后输出三位即可。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
#include<string>
using namespace std;
int main()
{
int n,i;
string a;
cin >> n;
cin >> a;
for (i = 0; i<n; i++)
{
cout << a[i];
if (i % 2 && n - i - 1>1)
{
cout << '-';
}
}
cout << endl;
return 0;
}
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