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

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

 1 #include <iostream>
 2 #include <string>
 3 using namespace std;
 4 string A, B;//注意，10^100超出double的范围，只能使用字符串来存取
 5 int N;
 6 int dealString(string &str)//数据预处理,返回数据位数
 7 {
 8     int k = str.find('.');//找到小数点
 9     if (k != -1)
10     {
11         str.erase(k, 1);//删除小数点
12         if (str[0] == '0')
13         {
14             k = 0;
15             str.erase(0, 1);//删除第一个0
16         }
17     }
18     else//没有小数
19     {
20         if (str != "0")
21             k = str.length();
22         else
23         {
24             k = 0;
25             str.erase(0, 1);//删除第一个0
26         }
27     }
28     while (!str.empty() && str[0] == '0')
29     {
30         str.erase(0, 1);//输出前面的0
31         k--;//比如0.000128 = 0.128*10^-3
32     }
33     if (str.empty())//这个数就是0
34         k = 0;
35     while (str.length() < N)
36         str += "0";//位数不够0来凑
37     return k;
38 }
39 
40 int main()
41 {
42     cin >> N >> A >> B;
43     //使用k1,k2来得到A,B的位数
44     int k1, k2;
45     k1 = dealString(A);
46     k2 = dealString(B);
47     A.assign(A.begin(), A.begin() + N);//取前N位
48     B.assign(B.begin(), B.begin() + N);
49     if (A == B && k1 == k2)
50     {
51         cout << "YES ";
52         cout << "0." << A << "*10^" << k1 << endl;
53     }
54     else
55     {
56         cout << "NO ";
57         cout << "0." << A << "*10^" << k1 << " ";
58         cout << "0." << B << "*10^" << k2 << endl;
59     }
60     return 0;    
61 }