Educational Codeforces Round 14

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

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#define inf 1<<30
#define maxn 10005
#define LOOP(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
string ch;
void clear_first_zero()
{
string::iterator it = ch.begin();
if (*it == '.'){
ch.insert(it, '0');
return;
}
int t=0;
for (int i = 0; i < ch.size(); i++)
{
if (ch[i] == '0'&&ch[i+1]!='.')
t++;
else
break;
}
ch = ch.erase(0, t);
if (!ch.size())
ch = "0";
}
void clear_last_zero()
{
int pos = ch.find('.', 0);
if (pos == -1)
{
//ch += '.';
return;
}
int t = ch.size();
for (int i = ch.size() - 1; i >= 0; i--)
{
if (ch[i] == '0') t--;
else if (ch[i] == '.') { t--; break; }
else break;
}
ch = ch.erase(t, ch.size());
}
int move()
{
int pos = ch.find('.', 0), change = 0;
if (ch.size() == 1) return 0;
if (pos == -1 ) {
ch += '.';
pos = ch.find('.', 0);
}
if (pos >= 2)
{
for (int i = pos; i >= 2; i--)
{
swap(ch[i], ch[i - 1]);
//Right = sum(Right, add);
change++;
}
//clear_first_zero();
clear_last_zero();
}
else
{

string::iterator it = ch.begin();
if (*it != '0'&&*(it+1)=='.'&&(it+2)!=ch.end()||ch=="0") return 0;
for (int i = 0; i < ch.size(); i++)
{
if (ch[i] != '0'&& ch[i] != '.')
break;
if (ch[i] != '.') change--;
}
ch=ch.erase(0, -change+1);
if(ch.size()>1)
ch.insert(ch.begin() + 1, '.');
}
return change;
}
int main()
{
//freopen("Text.txt", "r", stdin);
cin >> ch;
clear_first_zero();
clear_last_zero();
int t = move();
if (t > 0)
cout << ch << "E" << t << endl;
else if (t < 0)
else
cout << ch << endl;
}

大神的代码：
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#define inf 1<<30
#define maxn 1000005
#define LOOP(a,b,c) for(int a=b;a<=c;a++)
using namespace std;
int main()
{
string s;
cin >> s;
int pos = s.find('.');
if (pos != -1) {
s.erase(pos, 1);
} else {
pos = s.size();
}
int pos2 = 0;
for (int i=0; i<s.size(); i++) {
if (s[i] > '0') {
pos2 = i;
break;
}
}
int e = pos - pos2 - 1;
s.erase(0, pos2);
while (s.size() > 1 && s.back() == '0') {
s.pop_back();
}
if (s.size() >= 2) {
s.insert(1, ".");
}
if (e == 0) {
cout << s << endl;
} else {
cout << s << "E" << e << endl;
}
}