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

分析

此处考察字符串的处理
注：本题在第一次提交时测试点4（从0开始计数）一直通不过，最后再查看if语句的边界条件是发现边界判断条件出错，测试点4测试的是当指数为正整数时，此时要考虑若原有的小数位位数比指数小则要考虑在尾部追加0，然后再确定小数点’’.'的位置，测试点4考察当小数位数大于指数，小数点位数是否正确。在做条件判断时候，一定要考虑清楚边界条件，这个往往是测试点。

#include <iostream>
#include <cmath>
using namespace std;
int main() {
	string s;
	cin>>s;
	int flag=s[0]=='-'?-1:1;
	s.erase(0,1);
	int k=2; 
	while(k<(int)s.length() && s[k]!='E') k++;
	int flag2=s[k+1]=='-'?-1:1;
	string c=s.substr(0,k).erase(1,1),e=s.substr(k+2,s.length());
	int e_num=0;
	for(int i=0; i<(int)e.length(); i++) {
		e_num += (e[i]-'0')*pow(10,e.length()-i-1);
	}
	if(e_num!=0) {
		if(flag2<0) {
			for(int i=0; i<e_num; i++)
				c.insert(0,"0");
			c.insert(1,".");
		} else if(flag2>0) {
			int len_c=c.length();
			for(int i=0; i<e_num-len_c+1; i++) {
				c.push_back('0');
			}
			if(e_num<len_c-1) c.insert(e_num+1,".");
		}
	}else{
		c.insert(1,".");
	}
	if(flag<0) c.insert(0,"-");
	cout<<c;
	return 0;
}