Codeforces-691C-Exponential notation（模拟）

C. Exponential notation

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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 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

注意去除前导零和后导零

代码

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<queue>
#include<iomanip>
using namespace std;
const int maxn=1e6+5;
char str1[maxn];//接收数据
char str2[maxn];//存储处理好的结果
int main()
{
scanf("%s",str1);
int len_str1=strlen(str1);
int point=len_str1;//记录小数点指针
for(int i=0; i<len_str1; i++)//先找到小数点的位置
if(str1[i]=='.')
{
point=i;
break;
}
int flag_point;//寻找E后面的数字
int len_str2;//结果的指针
for(int i=0; i<len_str1; i++)//获取结果
{
if(str1[i]>'0'&&str1[i]<='9')
{
str2[0]=str1[i];
str2[1]='.';
flag_point=i+1;
len_str2=1;
for(int j=i+1; j<len_str1; j++)
if(str1[j]>='0'&&str1[j]<='9')
str2[++len_str2]=str1[j];
break;
}
}
for(int i=len_str2; i>=1; i--)//去除结果中的末尾零
{
if(str2[i]=='0'||str2[i]=='.')
len_str2--;
else
break;
}
for(int i=0; i<=len_str2; i++) //输出E前面的结果
printf("%c",str2[i]);
if(point!=flag_point)//如果需要输出E后面的数
point>flag_point?printf("E%d\n",point-flag_point):printf("E%d\n",point-flag_point+1);
return 0;
}