PAT 1073 Scientific Notation[字符串处理][科学记数法]

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1073 Scientific Notation（20 分）

Scientific notation is the way that scientists easily handle very large numbers or very small numbers. The notation matches the regular expression [+-][1-9].[0-9]+E[+-][0-9]+ which means that the integer portion has exactly one digit, there is at least one digit in the fractional portion, and the number and its exponent‘s signs are always provided even when they are positive.

Now given a real number

Input Specification:

Each input contains one test case. For each case, there is one line containing the real number

Output Specification:

For each test case, print in one line the input number

Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

-1.2E+10

Sample Output 2:

-12000000000

题目大意：将科学记数法转换为正常表示的数。由于给的范围都比较大，肯定只能用字符串来表示了

//之前见过一道科学记数法的题目呢，最关键的就是小数点的位置和首位非0元的位置。

pat上一次就AC，牛客网上也通过了，开心。

#include <iostream>
#include<stdlib.h>
#include <string>
#include<cmath>
using namespace std;

int main()
{
    string s;
    cin>>s;
    //找到指数。
    int pos=s.find("E");
    int expo=atoi(s.substr(pos+1).c_str());
    //如果指数是负数，那么就往前加0；
    if(expo<0){
        int p=s.find(".");
        s.erase(p,1);
        s.insert(1,"0.");
        int to=abs(expo)+2;
        for(int i=3;i<to;i++){
            s.insert(i,"0");
        }
        s=s.substr(0,pos+abs(expo));
        //cout<<s;

    }else if(expo>0){
        //找到小数点后有几位
        int len=pos-3;
       // cout<<len<<‘\n‘;
        if(len>expo){
            s.insert(expo+3,".");
            s.erase(2,1);
            s=s.substr(0,pos);
        }else{//len<=expo
            int p=expo-len;
            s=s.substr(0,pos);
            for(int i=0;i<p;i++){
                s+="0";//怎么把0放进去呢？
            }
            s.erase(2,1);
        }

    }
    if(s[0]==‘+‘)
        s=s.substr(1);
    cout<<s;
    return 0;
}
//  +1.2345E03
//  -1.2E+10