codeforces 691C C. Exponential notation(科学计数法)

题目链接：

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·10b, 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 aand 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

题意：

把给出的这个数用科学计数法法表示;

思路：

注意前导0,末尾的0,还有就是小数点的位置等;

AC代码：

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

//const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e6+10;
const int maxn=1005;
const double eps=1e-10;


char s[N],ans[N];

int main()
{
      scanf("%s",s);

      int len=strlen(s),pos=len,lastpos,cnt;
      For(i,0,len-1)
      {
        if(s[i]=='.')
        {
          pos=i;
          break;
        }
      }
      //if(s[0]=='.')pos=0;
      For(i,0,len-1)
      {
        if(s[i]>'0'&&s[i]<='9')
        {
          ans[0]=s[i];
          ans[1]='.';
          lastpos=i+1;
          cnt=1;
          for(int j=i+1;j<len;j++)
          {
            if(s[j]>='0'&&s[j]<='9')ans[++cnt]=s[j];
          }
          break;
        }
      }
      for(int i=cnt ;i>=1;i--)
      {
        if(ans[i]=='0'||ans[i]=='.')cnt--;
        else break;
      }
      for(int i=0;i<=cnt;i++)
      {
        printf("%c",ans[i]);
      }
      if(pos!=lastpos)
      {
        if(pos>lastpos)
        {
          printf("E%d",pos-lastpos);
        }
        else
        {
          printf("E%d",pos-lastpos+1);
        }
      }
      printf("
");
        return 0;
}