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.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

按整数部分是否为 0 来分情况讨论，求出本体和指数，然后判断两者是否都相等：

1. 整数部分为 0：由于在小数点后面还可能跟着若干个 0，因此本体部分是从小数点后第一个非零位开始的部分（不足 n 位则补 0），指数则是小数点与该非零位之间的 0 的个数的相反数。
2. 整数部分不为 0：本体部分是去掉小数点后的第一个非零位开始的 n 个字符（不足 n 位则补 0），指数则是小数点或最后一位（如没有小数点）之前的非零位数。

不管是哪种情况，如果有小数点，都需要把小数点去掉，这样才能输出正确的本体。

注意点

• 此题有一个陷阱：数据有可能出现前导 0，应该先去除前导零，然后再根据首位是否为小数点来判断是情况 1，还是情况2。
• 去除前导零后，若字符串长度为 0，则代表该数为 0，应将指数值为 0，否则有可能因小数点后全为 0 的情况，指数变为负数。

#include <iostream>
#include <string>
using namespace std;

int n;
string a, b;

void chopping(string &str, int &e) {
    while (str[0] == '0' && str.length() > 0) {
        str.erase(0, 1);
    }
    if (str[0] == '.') {
        str.erase(0, 1);
        while (str[0] == '0' && str.length() > 0) {
            str.erase(0, 1);
            e--;
        }
    } else {
        if (str.find('.') == string::npos) {
            e = str.length();
        } else {
            e = str.find('.');
            str.erase(str.find('.'), 1);
        }
    }
    int len = str.length();
    if (len == 0) {
        e = 0;
    }
    if (str.length() < n) {
        for (int i = 0; i < n - len; i++) {
            str.append("0");
        }
    } else if (len > n) {
        str = str.substr(0, n);
    }
}

int main() {
    cin >> n >> a >> b;
    int e1 = 0, e2 = 0;
    chopping(a, e1);
    chopping(b, e2);
    if (a == b && e1 == e2) {
        printf("YES");
        cout << " 0." << a << "*10^" << e1;
    } else {
        printf("NO");
        cout << " 0." << a << "*10^" << e1;
        cout << " 0." << b << "*10^" << e2;
    }
    return 0;
}