PAT甲级1073 Scientific Notation (20 分)

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 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

      题目大意：

         给出一个科学计数法，求原数

 

思路：

         分析：n保存E后面的字符串所对应的数字，t保存E前面的字符串，不包括符号位。当n<0时表示向前移动，那么先输出0. 然后输出abs(n)-1个0，然后继续输出t中的所有数字；当n>0时候表示向后移动，那么先输出第一个字符，然后将t中尽可能输出n个字符，如果t已经输出到最后一个字符(j == t.length())那么就在后面补n-cnt个0，否则就补充一个小数点。 然后继续输出t剩余的没有输出的字符～

     来源：https://blog.youkuaiyun.com/liuchuo/article/details/52121350

参考代码：

#include <iostream>
using namespace std;
int main() {
    string s;
    cin >> s;
    int i = 0;
    while (s[i] != 'E') i++;
    string t = s.substr(1, i - 1);
    int n = stoi(s.substr(i + 1));
    if (s[0] == '-') cout << "-";
    if (n < 0) {
        cout << "0.";
        for (int j = 0; j < abs(n) - 1; j++) cout << '0';
        for (int j = 0; j < t.length(); j++)
            if (t[j] != '.') cout << t[j];
    } else {
        cout << t[0];
        int cnt, j;
        for (j = 2, cnt = 0; j < t.length() && cnt < n; j++, cnt++) cout << t[j];
        if (j == t.length()) {
            for (int k = 0; k < n - cnt; k++) cout << '0';
        } else {
            cout << '.';
            for (int k = j; k < t.length(); k++) cout << t[k];
        }
    }
    return 0;
}

代码参考：https://blog.youkuaiyun.com/liuchuo/article/details/52121350