CodeForces 691C Exponential notation

题目：

Description

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 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.

Sample Input

Input

16

Output

1.6E1

Input

01.23400

Output

1.234

Input

.100

Output

1E-1

Input

100.

Output

1E2

因为x是大于0的，所以很容易求得最左边的和最右边的非0数字，不会出现什么意外。

求小数点的位置的时候，如果不存在小数点，就假设在末尾增加一个小数点，也就是dot初始化为strlen(c)

最后输出E后面的内容的时候还挺有意思的，2种情况表达式不一样。

代码：

#include<iostream>

using namespace std;

char c[1000002];
int main()
{
cin.getline(c, 1000002);
int l = strlen(c);
int start = -1, end = -1;
int dot = l;
for (int i = 0; i < l; i++)
{
if (c[i] == '.')dot = i;
else if (c[i] != '0')
{
if (start < 0)start = i;
end = i;
}
}
cout << c[start];
if (start != end)
{
cout << ".";
for (int i = start + 1; i <= end; i++)
{
if (i != dot)cout << c[i];
}
}
if (dot>start + 1)cout << "E" << dot - start - 1;
if (dot < start)cout << "E" << dot - start;
return 0;
}