Pat Problem Solving report [1073]

Tag: String

1073. Scientific notation (20) Time Limit 100 MS
The memory limit is 32000 kb.
Code length limit: 16000 B
Criterion author Hou, Qiming

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 file 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

 

The essence of this question is to simulate the transformation of scientific notation. In this sentence, the number is no more than 9999 bytes in length. What does this mean? The number of digits can be very long! At the beginning, I did not pay attention to this sentence. I converted the string into numbers for computation. The last two points of the result failed. I looked at it carefully and found this sentence in the question. Changed to the algorithm used to simulate the computation process of the string, AC.

Code:

// Pat-1073.cpp: defines the entry point for the console application. // # Include "stdafx. H "# include" iostream "# include" string "# include" stdio. H "# include" vector "# include" fstream "using namespace STD; vector <char> ans_pre; vector <char> ans_apend; int main () {// + 1.23400e-03 // Shishu: 1.23400 zhishu: 03 // OP: + dir:-fstream Fcin; Fcin. open ("C :\\ Users \ Administrator \ Desktop \ 456.txt"); string STR; Fcin> STR; int offset = Str. find ('E'); string Shishu = Str. substr (1, offset-1); // real number part int s Hishu_l = Shishu. length (); // the length of a real number. Char op = STR [0]; char dir = STR [Offset + 1]; int offset2 = Str. length ()-1-offset-1; string expart = Str. substr (Offset + 2, offset2); double zhishu = atof (expart. c_str (); int apend = 0; int Pos = 0; // original decimal place if (DIR = '-') pos-= zhishu; elsepos + = zhishu; if (Pos <0) {While (Pos <0) {ans_pre.push_back ('0'); POS ++ ;}} else {apend = pos-(shishu_l-2 ); // if the value is smaller than 0, the while (apend> 0) {ans_apend.push_back ('0'); APEN D --;} int Index = Shishu. find ('. '); Shishu. erase (index, 1); If (apend <0) // move the decimal point {Shishu. insert (shishu_l-1 + apend ,". ");} If (OP = '-') {cout <'-';} If (! Ans_pre.empty () {cout <ans_pre.back (); ans_pre.pop_back (); cout <"."; while (! Ans_pre.empty () {cout <ans_pre.back (); ans_pre.pop_back () ;}} cout <Shishu; while (! Ans_apend.empty () {cout <ans_apend.back (); ans_apend.pop_back () ;}return 0 ;}

Some string processing functions in this question can be carefully studied.