PAT甲级——A1060 Are They Equal

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 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 (<) 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 1, 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

其实这道题的难度在于得到这个数的幂次

 #include <iostream>
 #include <string>
 using namespace std;
 string A, B;//注意，10^100超出double的范围，只能使用字符串来存取
 int N;
 int dealString(string &str)//数据预处理,返回数据位数
 {
     int k = str.find('.');//找到小数点
     if (k != -)
     {
         str.erase(k, );//删除小数点
         if (str[] == '')
         {
             k = ;
             str.erase(, );//删除第一个0
         }
     }
     else//没有小数
     {
         if (str != "")
             k = str.length();
         else
         {
             k = ;
             str.erase(, );//删除第一个0
         }
     }
     while (!str.empty() && str[] == '')
     {
         str.erase(, );//输出前面的0
         k--;//比如0.000128 = 0.128*10^-3
     }
     if (str.empty())//这个数就是0
         k = ;
     while (str.length() < N)
         str += "";//位数不够0来凑
     return k;
 }

 int main()
 {
     cin >> N >> A >> B;
     //使用k1,k2来得到A,B的位数
     int k1, k2;
     k1 = dealString(A);
     k2 = dealString(B);
     A.assign(A.begin(), A.begin() + N);//取前N位
     B.assign(B.begin(), B.begin() + N);
     if (A == B && k1 == k2)
     {
         cout << "YES ";
         cout << "0." << A << "*10^" << k1 << endl;
     }
     else
     {
         cout << "NO ";
         cout << "0." << A << "*10^" << k1 << " ";
         cout << "0." << B << "*10^" << k2 << endl;
     }
     return ;
 }