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

17/25

 复制代码
#include <iostream>
#include <string>
#include <algorithm>
#include<math.h>
#include<queue>
#include<vector>
#include<set>
using namespace std;
 
int N;//精度
string A;
string B;
 
bool is_zero(string a) {//判断0
    if (a.length() == 1 && a[0] == 0) {
        return 1;
    }
    return 0;
}
 
string to_form(int &index, string a) {//这里不考虑a=0
    //int index = 0;//记录指数
    int len = a.length();
    string str;//记录去0后的数
    if (a[0] == '0') {//处理0.0000000...这类数
        for (int i = 0; i < len; i++) {
            if (a[i] >'0'&&a[i]<'9') {
                index = -(i - 2);
                str = a.substr(i, len - i);
                break;
            }
        }
    }
    else {//处理123456.789......和12345600......
        for (int i = 1; i < len; i++) {
            if (a[i] == '.') {
                index = i;
                str = a.substr(0, 5) + a.substr(i + 1, len - i + 1);
                break;
            }
            else if(i==len-1){//最后一位肯定不是小数点，所有只有整数才能遍历到最后一位
                index = len;
                str = a;
            }
        }
    }
    return "0." + str;
}
 
string deal(string a, int n,int &index) {//输出格式化
    string str;
    if (is_zero(a) == true) {
        for (int i = 0; i < n; i++) {
            str += '0';
        }
    str = "0." + str;
    return str;
    }
    else {
        str = to_form(index, a);
        if (str.length() < (n + 2)) {//位数不足后面补零
            for (int i = 0; i < (n + 2 - str.length()); i++) {
                str = str + '0';
            }
        }
        if (str.length() > (n + 2)) {//位数过长截尾
            str.erase(n + 2, str.length() - (n + 2));
        }
        return str;
   }
}
 
int main() {
    cin >> N;
    cin >> A;
    cin >> B;
    int index_a = 0;
    int index_b = 0;//a,b指数
    string s_a;
    string s_b;
    s_a = deal(A,N,index_a);
    s_b = deal(B, N,index_b);
    //cout << s_a << "*10^" << index_a;
    if (s_a == s_b&&index_a==index_b) {
        cout << "YES " << s_a<<"*10^"<<index_a;
    }
    else {
        cout << "NO " << s_a <<"*10^"<<index_a<< ' ' << s_b<<"*10^"<<index_b;
    }
}