【PAT】1060 Are They Equal (25)（25 分）

1060 Are They Equal (25)（25 分）

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

 

科学计数法，注意以下几种情况：

　　1.数字前有 0 的，比如：00056；0000.32 等，需先去除无效的0；

　　2.指数 e 的正负计算,0.1 = 0.1*10^0;

　　3.不同情况下有效位的计算，比如00056，0.00032；

　　4.按题目要求输出有效位数，不足的后面补0；

 

C++代码如下：

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


void deal(string &s, int &e) {
    while (s.length() > 0 && s[0] == '0') s.erase(s.begin());
    int i;
    if (s[0] == '.') {
        s.erase(s.begin());
        while (s.length()>0&&s[0] == '0') {
            e--; 
            s.erase(s.begin());
        }
        if (s.length() == 0) e = 0;
    }
    else {
        for ( i = 0; i < s.length(); i++) {
            if (s[i] != '.')    e++;                
            else break;
        }
        if(i<s.length()) s.erase(s.begin() + i);
    }
}
void signdigit(string &s,int n) {
    while (s.length() < n) s += '0';
    s.resize(n);
}
int main() {
    string str1, str2;
    int n,e1=0,e2=0;
    cin >> n >> str1 >> str2;
    deal(str1, e1);
    deal(str2, e2);
    signdigit(str1, n);
    signdigit(str2, n);
    if (str1 == str2 && e1 == e2)
        cout << "YES 0." << str1 << "*10^" << e1 << endl;
    else
        cout << "NO 0." << str1 << "*10^" << e1 << " 0." << str2 << "*10^" << e2 << endl;
    return 0;
}