A1060.Are They Equal

浮点数等价判断

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本文介绍了一种算法，用于判断两个浮点数在给定有效位数的机器上是否被视为相等。输入包括有效位数及两个待比较的非负浮点数，输出则表明这两个数是否等价，并给出其标准形式。

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*105 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 10100, 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.d1…dN*10^k” (d1>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

Code：

#include "iostream"
#include "string"

using namespace std;

int n = 0;   //有效位
string deal(string s,int& e)
{
    int k = 0;  //s的小标
    while(s.length() > 0 && s[0] == '0')
    {
        s.erase(s.begin());   //去掉s的前导零
    }

    if(s[0] == '.')  //去掉前导零后出现小数点，说明s是小于1的小数
    {
        s.erase(s.begin());   //去掉小数点
        while(s.length() > 0 && s[0] == '0')
        {
            s.erase(s.begin());   //去掉小数点后非零位前的所有零
            e--;    //去掉一个0，指数e减1
        }
    }
    else  //去掉前导零后不是小数点，则找到后面的小数点删除
    {
        while(k < s.length() && s[k] != '.')  //寻找小数点
        {
            k++;
            e++;  //只要不碰到小数点就让指数e++
        }
        if(k < s.length()) //碰到小数点
        {
            s.erase(s.begin() + k);
        }
    }

    if(s.length() == 0)  //如果去掉前导零后s长度为0，说明这个数为0
    {
        e = 0;
    }

    int num = 0;
    k = 0;
    string res;
    while(num < n)  //只要精度还没到n
    {
        if(k < s.length()) res += s[k++]; //还有数字就加到res末尾
        else res += '0';
        num ++;  //精度加1
    }
    return res;
}

int main()
{
    string s1,s2,s3,s4;
    cin >> n >> s1 >> s2;
    int e1 = 0, e2 = 0;
    s3 = deal(s1,e1);
    s4 = deal(s2,e2);
    if(s3 == s4 && e1 == e2)
    {
        cout<<"YES 0."<<s3<<"*10^"<<e1<<endl;
    }
    else
    {
        cout<<"NO 0."<<s3<<"*10^"<<e1<<" 0."<<s4<<"*10^"<<e2<<endl;
    }

    return 0;
}