1060. Are They Equal (25)

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.d1...dN*10^k" (d1>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

IDEA

别人的AC代码 借鉴

1.需要将浮点数转成科学计数法，然后确定有效位数，进行比较

2.用到string的函数insert() 和 eraser()

CODE

#include<iostream>
#include<cstring>
#include<fstream>
using namespace std;

void machine(string *str,int *e,int N){
	int dot=(*str).find('.');
	if(dot==-1){//没找到小数点 
		(*e)=str->size();
	}else{
		str->erase(dot,1);//去掉小数点 
		(*e)=dot;
	}
	
	str->insert(0,"0.");
	
	dot=2;
	while(str->size()>dot&&(*str)[dot]=='0'){//小数点后边的0 
		dot++;
	} 
	dot-=2;
	str->erase(2,dot);//去掉小数点后边的0 
	(*e)-=dot;
	
	dot=str->size();
	if(dot==2){
		(*e)=0;
	}
	N+=2;
	if(dot>N){//如果数字本身长度大于有效数位 
		str->erase(N,dot-N+1);//把后边无效的去掉 
	}else{
		str->insert(dot,N-dot,'0');
	}
}
int main(){
	#ifndef ONLINE_JUDGE
	freopen("input.txt","r",stdin);
	#endif
	
	int N;
	string A,B;
	int A_e,B_e;
	cin>>N>>A>>B;
	
	machine(&A,&A_e,N);
	machine(&B,&B_e,N);
	
	if(A==B&&A_e==B_e){
		cout<<"YES "<<A<< "*10^" <<A_e<<endl;
	}else{
		cout<<"NO "<<A<< "*10^" <<A_e<<" "<<B<< "*10^" <<B_e<<endl;
	}
	
	#ifndef ONLINE_JUDGE
	fclose(stdin);
	#endif
	return 0;
}