PAT甲级——A1060 Are They Equal-白红宇

PAT甲级——A1060 Are They Equal

阅读量：4540 次

发布时间：2019-06-08

本文共 2627 字，大约阅读时间需要 8 分钟。

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 
     
       2 #include 
      
        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016         }17     }18     else//没有小数19     {20         if (str != "0")21             k = str.length();22         else23         {24             k = 0;25             str.erase(0, 1);//删除第一个026         }27     }28     while (!str.empty() && str[0] == '0')29     {30         str.erase(0, 1);//输出前面的031         k--;//比如0.000128 = 0.128*10^-332     }33     if (str.empty())//这个数就是034         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     else55     {56         cout << "NO ";57         cout << "0." << A << "*10^" << k1 << " ";58         cout << "0." << B << "*10^" << k2 << endl;59     }60     return 0;    61 }

转载于:https://www.cnblogs.com/zzw1024/p/11294212.html

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