PAT 1073. Scientific notation (20)

http://pat.zju.edu.cn/contests/pat-a-practise/1073

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

Now given a real number a in scientific notation, and you are supposed to print a in the conventional notation while keeping a ll the significant figures.

Input Specification:

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

Output Specification:

For each test case, print at one line the input number A in the conventional notation and the significant figures ke PT, including trailing zeros, Sample Input 1:

+1.23400E-03

Sample Output 1:

0.00123400

Sample Input 2:

-1.2E+10

Sample Output 2:

-12000000000

Malle an Egg ~ positive number to go ' + ', did not notice that this wasted too much time ~ ~

#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
using namespace std;

int main () {
	string s, ans = "";
	Getline (CIN, s);
	if (s[0] = = '-')
		ans + = s[0];
	int indexe = S.find ("E");
	String num = S.substr (1, indexE-1);
	char x = S[indexe + 1];
	String exp = S.substr (indexe+2, S.size ()-indexE-2);
	StringStream SS;
	SS << Exp;
	int e;
	SS >> E;
	if (E = = 0) {
		cout << ans << num << endl;
		return 0;
	}
	if (x = = ' + ') {
		if (E < Num.size ()-2) {
			ans = ans + num[0] + num.substr (2, E) + "." + num.substr (E + 2, NUM.S Ize ()-e-2);
		else{
			ans = ans + num[0] + num.substr (2, Num.size ()-2);
			for (int i = 0; i < e-num.size () + 2; i++)
				ans + = "0";
		}
	}
	if (x = = '-') {
		ans = ans + "0.";
		while (e--! = 1)
			ans + = "0";
		Ans = ans + num[0] + num.substr (2, Num.size ()-2);
	}
	cout << ans << endl;
	return 0;
}