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
代码
#include<iostream>
#include<string>
#include<stdio.h>
#define maxn 105using namespace std;string transfer(int n,string s){int i=0,num[105]={0},flag=0,start,dot,notfound=0;string result;for(auto it=s.begin();it!=s.end();it++){if((*it!='0'||flag)&&*it!='.'){flag=1;if(i==0) start=it-s.begin();num[i++]=*it-'0';//printf("%d\n",num[i-1]);}if(i==n)break; }if(i==0)notfound=1;if(s.find('.')!=-1) dot=s.find('.');else dot=s.length();result="0.";for(int j=0;j<n;j++)result+=std::to_string(num[j]);//cout<<result<<endl;result+="*10^";if(notfound)result+="0";else if(dot-start>0)result+=std::to_string(dot-start);elseresult+=std::to_string(dot-start+1);return result;
}int main(){int n;string a,b;string str1,str2;//scanf("%d %s %s",&n,a,b);cin>>n>>a>>b;str1=transfer(n,a);str2=transfer(n,b);if(str1!=str2)cout<<"NO "<<str1<<" "<<str2<<endl; elsecout<<"YES "<<str1<<endl;return 0;
}