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
参考代码：
#include<iostream>
#include<string>
using namespace std;
int n;
string convert(const string &s, int & exp)
{
int cnt = s.size(), p = 0;
for (int i = 0; i < s.size(); i++) {
if (s[i] == '.') {
cnt = i;
break;
}
}
while (s[p] == '0' || s[p] == '.')p++;
if (cnt>=p) {
cnt=cnt-p;
}
else cnt = cnt - p + 1;
if (p == s.size())cnt = 0;
int index = 0;
string num;
while (index<n) {
if (p < s.size() && s[p] != '.')
{
num += s[p];
index++;
}
else if (p >= s.size())
{
num += '0';
index++;
}
p++;
}
exp = cnt;
return num;
}
int main()
{
string a, b;
cin >> n >> a >> b;
string sa, sb;
int na, nb;
sa = convert(a, na);
sb = convert(b, nb);
if (sa == sb && na == nb) {
cout << "YES 0." << sa << "*10^" << na;
}
else cout << "NO " << "0." << sa << "*10^" << na << " 0." << sb << "*10^" << nb;
return 0;
}