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 0.12010^3 0.13810^3
代码：
#include<iostream>
#include<string>
using namespace std;
int n;
string deal(string s, int &e) {
int k=0;
while (s.length()>0&&s[0]=='0')
{
s.erase(s.begin());
}if (s[0]=='.')
{
s.erase(s.begin());
while (s.length()>0&&s[0]=='0')
{
s.erase(s.begin());
e--;
}
}
else
{
while (s.length()>k&&s[k]!='.')
{
k++;
e++;
}
if (k<s.length())
{
s.erase(s.begin()+k);
}
}
if (s.length()==0)
{
e = 0;
}
string res;
int num = 0;
k = 0;
while (num<n)
{
if (k<s.length())
{
res += s[k++];
}
else res += '0';
num++;
}
return res;
}
int main(){
string s1, s2, s3, s4;
int e1 = 0, e2 = 0;
cin >> n >> s1 >> s2;
s3 = deal(s1, e1);
s4 = deal(s2, e2);
if (s3==s4&&e1==e2)
{
cout << "YES 0." << s3 << "*10^" << e1<<endl;
}
else
{
cout << "NO 0." << s3 << "*10^" << e1<< " 0." << s4 << "*10^" << e2 << endl;
}
return 0;
}
参考晴神的《算法笔记》
