Codeforces 691C. Exponential notation 模拟题

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[1000100];
int main()
{
    int i;
    scanf("%s",x);
    int len=strlen(x);
    int s=-1,e=len-1,sign=len;
    for(i=0; i<len; i++)
        if(x[i]=='.')
        {
            sign=i;
            break;
        }
    for(i=0; i<len; i++)
        if(x[i]>'0'&&x[i]<='9')
        {
            s=i;
            break;
        }
    for(i=len-1; i>=0; i--)
        if(x[i]>'0'&&x[i]<='9')
        {
            e=i;
            break;
        }
    if(s>=0)
    {
        cout<<x[s];
        if(e>s) cout<<".";
        for(i=s+1; i<=e; i++)
            if(x[i]!='.') cout<<x[i];
        if((s+1)!=sign)
        {
            cout<<"E";
            if(s<sign) cout<<sign-s-1<<endl;
            else if(s>sign) cout<<sign-s<<endl;
        }
    }
    else cout<<"0"<<endl;
    return 0;
}

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转载于:https://www.cnblogs.com/GeekZRF/p/5687860.html