pat (advanced level) practice 1073 scientific notation (详细注释）_joker_sxj的博客-爱代码爱编程

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 A in scientific notation, you are supposed to print A in the conventional notation while keeping all the significant figures.

Input Specification:

Each input contains one test case. For each case, there is one line containing the real number A in scientific notation. The number is no more than 9999 bytes in length and the exponent's absolute value is no more than 9999.

Output Specification:

For each test case, print in one line the input number A in the conventional notation, with all the significant figures kept, including trailing zeros.

Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

-1.2E+10

Sample Output 2:

-12000000000

#include<bits/stdc++.h>
using namespace std;
int main()
{
    string s;   cin >> s;
    int i = 0;
    while(s[i] != 'E')  i++;    //i存储E的位置
    string t = s.substr(1,i-1); //t存储E前面的不含符号的数字
    int n = stoi(s.substr(i+1));//n存储E后面带符号的幂数
    if(s[0] == '-') cout << "-";
    if(n < 0){                  //如果幂数小于0
        cout << "0.";           //因为科学计数法前面的数字<10,故n<0必有“0.”作开头
        for(int j = 0; j < abs(n) - 1; ++j)     cout << '0';//输出中间n-1个0，注意输出单个字符用‘’，多个字符用“”
        for(int j = 0; j < t.length(); ++j)
            if(t[j] != '.') cout << t[j];    //输出幂数小于0时E前面除小数点之外需要输出的数字，避免了0的不输出情况，很好
    }else{
        cout << t[0];           //输出E前面的整数位
        int cnt, j;
        for(j = 2, cnt = 0; j < t.length() && cnt < n; ++j, ++cnt)  cout << t[j];
        //cnt存储幂，然后通过j输出E前面的小数部分
        if(j == t.length()){    //如果幂次大于小数点的位数，即j先结束循环，所以需要输出幂次还未输出的0
            for(int k = 0; k < n - cnt; ++k)    cout << '0';
        } else {                //如果cnt先结束循环，需要先输出小数点，然后继续输出t未输出的部分
            cout << '.';
            for(int k = j; k < t.length(); k++)     cout << t[k];
        }
    }
    return 0;
}