Coderfroces 691 C. Exponential notation

C. Exponential notation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a positive decimal number x.

Your task is to convert it to the "simple exponential notation".

Let x = a·10^b, where 1 ≤ a < 10, then in general case the "simple exponential notation" looks like "aEb". If b equals to zero, the part "Eb" should be skipped. If a is an integer, it should be written without decimal point. Also there should not be extra zeroes in a and b.

Input

The only line contains the positive decimal number x. The length of the line will not exceed 10⁶. Note that you are given too large number, so you can't use standard built-in data types "float", "double" and other.

Output

Print the only line — the "simple exponential notation" of the given number x.

Examples

Input

16

Output

1.6E1

Input

01.23400

Output

1.234

Input

.100

Output

1E-1

Input

100.

Output

1E2
模拟题，但是情况都点多，自己还是太弱，代码估计显得冗长

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <vector>
#include <iomanip>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
#define lowbit(x) (x&(-x))
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define MAX 100000000000000000
#define MOD 1000000007
#define pi acos(-1.0)
#define ei exp(1)
#define PI 3.141592653589793238462
#define INF 0x3f3f3f3f3f
#define mem(a) (memset(a,0,sizeof(a)))
typedef long long ll;
string sf,sl,b;
char a[1000006];
void solve(char *a)
{
    sf="";sl="";b="";
    bool flag=false;
    bool del=false;
    for(int i=0;a[i]!='\0';i++)
    {
        if(a[i]=='.')
        {
            del=true;
            break;
        }
        if(a[i]=='0' && !flag) continue;
        else 
        {
            sf+=a[i];
            flag=true;
        }
    }
    if(sf.size()>0)
    {
        flag=false;
        for(int i=sf.size()-1;i>=0;i--)
        {
            if(sf[i]=='0' && flag==false) continue;
            else
            {
                b+=sf[i];
                flag=true;
            }
        }
        reverse(b.begin(),b.end());
    }
    if(del==true)
    {
        int len=strlen(a);
        flag=false;
        for(int i=len-1;i>=0;i--)
        {
            if(a[i]=='.') break;
            if(a[i]=='0' && !flag) continue;
            else
            {
                sl+=a[i];
                flag=true;
            }
        }
        reverse(sl.begin(),sl.end());
    }
    if(sf.size()>0)
    {
        if(sf.size()==1 && sl.size()==0)
        {
            cout<<sf<<endl;
        }
        if(sf.size()==1 && sl.size()!=0)
        {
            cout<<sf<<"."<<sl<<endl;
        }
        if(sf.size()>1 && sl.size()==0)
        {
            cout<<sf[0];
            if(b.size()==1) cout<<"E"<<sf.size()-1<<endl;
            else
            {
                cout<<".";
                for(int i=1;i<b.size();i++)
                cout<<b[i];
                cout<<"E"<<sf.size()-1<<endl;
            }
        }
        if(sf.size()>1 && sl.size()!=0)
        {
            cout<<sf[0]<<".";
            for(int i=1;i<sf.size();i++)
                cout<<sf[i];
            cout<<sl<<"E"<<sf.size()-1<<endl;
        }
    }
    else
    {
        int pos=1;
        int k=sl.size(),vis;
        for(int i=0;i<k;i++)
        {
            if(sl[i]=='0') pos++;
            else
            {
                vis=i;
                break;
            }
        }
        if(k-1==vis)
        {
            cout<<sl[vis]<<"E-"<<pos<<endl;
        }
        else
        {
            cout<<sl[vis]<<".";
            for(int i=vis+1;i<k;i++)
                cout<<sl[i];
            cout<<"E-"<<pos<<endl;
        }
    }
}
int main()
{
    scanf("%s",a);
    solve(a);
    return 0;
}