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  • Codeforces 691C Exponential notation(模拟)

    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.

    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<cstring>
    using namespace std;
    const int maxn=1e6+5;
    char s[maxn];
    int main()
    {
        while(cin>>s)
        {
            int len=strlen(s);
            int h=-1,t=len,p=len;//h第一个不为0的位置,t为最后一个不为0的位置,p是小数点的位置。
            for(int i=0;i<len;i++)
            {
                if(s[i]=='.')
                    p=i;
                if(s[i]!='.'&&s[i]!='0'&&h==-1)
                    h=i;
            }
            for(int i=len-1;i>=0;i--)
            {
                if(s[i]!='.'&&s[i]!='0')
                {
                    t=i;
                    break;
                }
            }
            if(h==-1)
                cout<<'0'<<endl;//说明都是0000.0000形式的
            else
            {
                cout<<s[h];//输出第一个数字
                if(h!=t)//后面存在有效数字
                    cout<<'.';
                for(int i=h+1;i<=t;i++)
                    if(s[i]!='.')
                        cout<<s[i];
                int e;//计算幂次
                if(h>p)
                    e=p-h;
                else
                    e=p-h-1;
                if(e)
                    cout<<'E'<<e<<endl;
                else
                    cout<<endl;
            }
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/orion7/p/7295339.html
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