Code Forces 20A BerOS file system

A. BerOS file system

time limit per test
2 seconds

memory limit per test
64 megabytes

input
standard input

output
standard output

The new operating system BerOS has a nice feature. It is possible to use any number of characters '/' as a delimiter in path instead of one traditional '/'.
For example, strings //usr///local//nginx/sbin// and /usr/local/nginx///sbin are
equivalent. The character '/' (or some sequence of such characters) at the end of the path is required only in case of the path to the root directory, which
can be represented as single character '/'.

A path called normalized if it contains the smallest possible number of characters '/'.

Your task is to transform a given path to the normalized form.

Input

The first line of the input contains only lowercase Latin letters and character '/' — the path to some directory. All paths start with at least one character '/'.
The length of the given line is no more than 100 characters, it is not empty.

Output

The path in normalized form.

Examples

input

//usr///local//nginx/sbin

output

/usr/local/nginx/sbin

遇到多个////
只输出一个/

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
char a[105];
char b[105];
int main()
{
gets(a);
bool tag=0;
int len=strlen(a);
int cnt=0;
for(int i=0;i<len;i++)
{
if(a[i]!='/')
{
// cout<<a[i];
b[cnt++]=a[i];
tag=0;
}

else
{
if(!tag)
{
//cout<<a[i];
b[cnt++]=a[i];
tag=1;
}
}
}
if(b[0]!='/')
cout<<'/';
for(int i=0;i<cnt;i++)
{
if(i==cnt-1&&b[i]=='/'&&cnt!=1)
continue;
cout<<b[i];
}
cout<<endl;
return 0;
}