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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

C++：

/* @Date    : 2018-02-02 11:05:27 @Author  : 酸饺子 (changzheng300@foxmail.com) @Link    : https://github.com/SourDumplings @Version : $Id$*//*https://www.patest.cn/contests/pat-a-practise/1060 */#include 
   
    #include 
    
     #include 
     
      using namespace std;string toChop(const string &num, const int prec){    int valid = 0;    int exp = 0;    bool decimal = false;    int n = num.length();    string result("0.");    int count = 0;    for (; valid != n && (num[valid] == '0' || num[valid] == '.'); ++valid)    {        if (decimal)            --exp;        if (num[valid] == '.')            // 说明这个数是个小于1的数，即纯小数            decimal = true;    }    if (valid == n)        // 说明这个数为0        exp = 0;    // printf("valid = %d, exp = %d\n", valid, exp);    for (int i = valid; i != n; ++i)    {        if (num[i] != '.')        {            if (count != prec)            {                result += num[i];                ++count;            }        }        else if (!decimal)        {            // 这个数不是纯小数并且整数部分读取完毕            exp = i - valid;            decimal = true;        }    }    if (exp == 0 && valid != n)        // 说明是个非零正整数        exp = n - valid;    if (count != prec)        result += string((prec - count), '0');    string exp_s = to_string(exp);    result += "*10^";    result += exp_s;    return result;}int main(int argc, char const *argv[]){    int N;    string s1, s2;    cin >> N >> s1 >> s2;    string result1 = toChop(s1, N), result2 = toChop(s2, N);    if (result1 == result2)    {        printf("YES ");        cout << result1 << endl;    }    else    {        printf("NO ");        cout << result1 << " " << result2 << endl;    }    return 0;}
     
    
   

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