A1060 Are They Equal
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
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <algorithm>
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());//去掉s的前导零
}
if(s[0]=='.'){//小于1的数
s.erase(s.begin());
while(s.length()>0 && s[0]=='0'){
s.erase(s.begin());//去掉小数点后非零位前的所有零
e--;//每去掉一个0,指数e减1
}
}
else{//大于1的数
while(k<s.length() && s[k]!='.'){
k++;
e++;
}
if(k<s.length())//把小数点删除
s.erase(s.begin()+k);
}
if(s.length()==0)
e=0;//如果去除前导零后s的长度变为0,则说明这个数是0
int num=0;
k=0;
string res;
while(num<n){//精度不到n
if(k<s.length()) res+=s[k++];
else res+='0';
num++;
}
return res;
}
int main(){
string s1,s2,s3,s4;
cin>>n>>s1>>s2;
int e1=0,e2=0;
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;
}