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
代码如下:
#include <iostream>
#include <string>
using namespace std;
int n;
string convert2std(string s,int &e){
int k=0; //s的下标
//清除前导0
while(s.size()>0 && s[0]=='0'){
s.erase(s.begin());
}
//去除小数点和小数点后的0
if(s[0]=='.'){
s.erase(s.begin());
while(s.size()>0 && s[0]=='0'){
s.erase(s.begin());
e--; //小数点后有几个0就是负指数几
}
}else{
for(;k<s.size();k++){
if(s[k]=='.'){ //去除小数点
s.erase(s.begin()+k);
break;
}
else
e++;
}
}
//这样留下来不是0的字符串了 即a1a2a3a4...
if(s.size()==0){
e=0;
}
string ans;
if(n<=s.size()){
ans=s.substr(0,n);
}else{
ans=s;
for(int i=0;i<n-s.size();i++){
ans+='0';
}
}
return ans;
}
int main()
{
string num1,num2;
cin >> n >> num1 >> num2;
int e1=0,e2=0;
string s1=convert2std(num1,e1);
string s2=convert2std(num2,e2);
if(s1==s2 && e1==e2){
cout << "YES 0." << s1 << "*10^" << e1 << endl;
}else{
cout << "NO 0." << s1 << "*10^" << e1 << " 0." << s2 << "*10^" << e2 << endl;
}
return 0;
}