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  • 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;
    }
    View Code
    
    
    
    
    
    
     
    I am a slow walker,but I never walk backwards.
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  • 原文地址:https://www.cnblogs.com/GeekZRF/p/5687860.html
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