【PAT】1060 Are They Equal (25)（25 分）

1060 Are They Equal (25)（25 分）

If a machine can save only 3 significant digits, the float numbers 12300 and 12358.9 are considered equal since they are both saved as 0.123*10^5^ with simple chopping. Now given the number of significant digits on a machine and two float numbers, you are supposed to tell if they are treated equal in that machine.

Input Specification:

Each input file contains one test case which gives three numbers N, A and B, where N (<100) is the number of significant digits, and A and B are the two float numbers to be compared. Each float number is non-negative, no greater than 10^100^, and that its total digit number is less than 100.

Output Specification:

For each test case, print in a line "YES" if the two numbers are treated equal, and then the number in the standard form "0.d~1~...d~N~*10\^k" (d~1~>0 unless the number is 0); or "NO" if they are not treated equal, and then the two numbers in their standard form. All the terms must be separated by a space, with no extra space at the end of a line.

Note: Simple chopping is assumed without rounding.

Sample Input 1:

3 12300 12358.9

Sample Output 1:

YES 0.123*10^5

Sample Input 2:

3 120 128

Sample Output 2:

NO 0.120*10^3 0.128*10^3

科学计数法，注意以下几种情况：

　　1.数字前有 0 的，比如：00056；0000.32 等，需先去除无效的0；

　　2.指数 e 的正负计算,0.1 = 0.1*10^0;

　　3.不同情况下有效位的计算，比如00056，0.00032；

　　4.按题目要求输出有效位数，不足的后面补0；

C++代码如下：

 #include<iostream>
 #include<string>
 #include<iomanip>
 using namespace std;

 void deal(string &s, int &e) {
     while (s.length() >  && s[] == '') s.erase(s.begin());
     int i;
     if (s[] == '.') {
         s.erase(s.begin());
         while (s.length()>&&s[] == '') {
             e--;
             s.erase(s.begin());
         }
         if (s.length() == ) e = ;
     }
     else {
         for ( i = ; i < s.length(); i++) {
             if (s[i] != '.')    e++;
             else break;
         }
         if(i<s.length()) s.erase(s.begin() + i);
     }
 }
 void signdigit(string &s,int n) {
     while (s.length() < n) s += '';
     s.resize(n);
 }
 int main() {
     string str1, str2;
     int n,e1=,e2=;
     cin >> n >> str1 >> str2;
     deal(str1, e1);
     deal(str2, e2);
     signdigit(str1, n);
     signdigit(str2, n);
     if (str1 == str2 && e1 == e2)
         cout << "YES 0." << str1 << "*10^" << e1 << endl;
     else
         cout << "NO 0." << str1 << "*10^" << e1 << " 0." << str2 << "*10^" << e2 << endl;
     return ;
 }