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 题目链接：http://codeforces.com/problemset/problem/691/C
题意：就是把一个数转换成a*10^b（1≤a﹤10）形式，输出aEb。
思路：标记第一个不为零的数的位置作为起点s，标记最后一个不为零的数的位置作为终点e，标记小数点的位置sign，默认位置应该为len+1。根据三个位置进行输出。 代码：
#include<bits/stdc++.h>
using namespace std;
char x[];
int main()
{
int i;
scanf("%s",x);
int len=strlen(x);
int s=-,e=len-,sign=len;
for(i=; i<len; i++)
if(x[i]=='.')
{
sign=i;
break;
}
for(i=; i<len; i++)
if(x[i]>''&&x[i]<='')
{
s=i;
break;
}
for(i=len-; i>=; i--)
if(x[i]>''&&x[i]<='')
{
e=i;
break;
}
if(s>=)
{
cout<<x[s];
if(e>s) cout<<".";
for(i=s+; i<=e; i++)
if(x[i]!='.') cout<<x[i];
if((s+)!=sign)
{
cout<<"E";
if(s<sign) cout<<sign-s-<<endl;
else if(s>sign) cout<<sign-s<<endl;
}
}
else cout<<""<<endl;
return ;
}