cf 619c 科学记数法

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 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.

Educational Codeforces Round 14
Finished
Practice

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→ Last submissions

Submission	Time	Verdict
19198555	Jul/18/2016 04:52	Accepted
19198494	Jul/18/2016 04:47	Wrong answer on test 1

→ Problem tags

implementation

No tag edit access

<span style="font-size:18px;">#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#define M 1000005
char s[M],ss[M];
using namespace std;
int main()
{
    int i,k,j,m,k1,k2;
    while(~scanf("%s",&s))
    {
        m=strlen(s);
        for(i=0; i<m; i++)
        {
            if(s[i]=='.')//标记小数点的位置
            {
                k1=i;
                break;
            }
            else k1=m;
        }
         
        for(i=0; i<m; i++)
        {
            if(s[i]>'0'&&s[i]<='9')
            {
                ss[0]=s[i];
                ss[1]='.';
                k=1;
                k2=i+1;
                for(j=i+1; j<m; j++)
                {if(s[j]>='0'&&s[j]<='9')
                    ss[++k]=s[j];
                }
                break;
            }
        }
        for(i=k; i>=1; i--)//去掉多余的0和小数点
        {
            if(ss[i]=='0'||ss[i]=='.')
                k--;
            else break;
        }
        for(i=0; i<k+1; i++)
            printf("%c",ss[i]);
        if(k1>k2)
        {
            printf("E%d",k1-k2);
        }
        else if(k1<k2)
            printf("E%d",k1-k2+1);
        printf("\n");
    }
    return 0;
}
</span>

Educational Codeforces Round 14
Finished
Practice