【PAT A1060】 Are They Equal

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原文链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805413520719872

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该程序实现了一个浮点数比较的功能，考虑了简单截断的情况，不进行四舍五入。输入包括有效位数和两个浮点数，输出判断结果及标准化形式。在处理过程中，对前导零和小数点进行了特殊处理，确保了比较的准确性。

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

思路：参考《算法笔记》。关键在于对整数部分是否为0来分两种情况：① 0.a1a2a3…，② b1b2…bm.a1a2a3… 。若本体和指数均相等，结果才相等。注意对前导0的处理。

#include <iostream>
#include <string>

using namespace std;
int n;//有效位数

string deal(string s, int &e) {//注意指数e是引用
    //对前导0的处理
    int k = 0;//s的下标
    while (s.length() > 0 && s[0] == '0') {
        s.erase(s.begin());//去掉s的前导零
    }
    //对小数点的处理
    if (s[0] == '.') {//去掉前导零后是小数点，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所有的零
            e--;//每去掉一个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] != '.') {//找到小数点
            k++;//下标+1
            e++;//只要不碰到小数点就让指数e++
        }
        if (k < s.length()) {//满足条件则说明找到了小数点
            s.erase(s.begin() + k);//删除小数点
        }
    }
    //0的特殊处理
    if (s.length() == 0) {//如果去除前导零后s的长度变为0，说明这个数是0
        e = 0;
    }
    int num = 0;//当前精度
    k = 0;
    string res;
    while (num < n) {//只要精度还没到n
        if (k < s.length()) res += s[k++];//只要还有数字，就加到res末尾
        else res += '0';//否则res末尾添加0
        num++;//精度加1
    }
    return res;
}

int main() {
    string s1, s2, s3, s4;
    cin >> n >> s1 >> s2;
    int e1 = 0, e2 = 0;//e1，e2为s1，s2的指数
    s3 = deal(s1, e1);
    s4 = deal(s2, e2);
    if (s3 == s4 && e1 == e2) {//主体相同且指数相同则YES
        cout << "YES 0." << s3 << "*10^" << e1 << endl;
    } else {
        cout << "NO 0." << s3 << "*10^" << e1 << " 0." << s4 << "*10^" << e2 << endl;
    }
    return 0;
}