很明顯我寫得太冗余了。。

簡單說明一下吧。。我以為0就是輸出0這個(gè)點(diǎn)錯(cuò)了一個(gè)，0要輸出3 0 0要輸出0.000*10^0。。還有一個(gè)點(diǎn)是前導(dǎo)0.如00123,。。這個(gè)點(diǎn)我真沒想到。。

其他點(diǎn)情況就很簡單了。。分成小于1和大于1處理就能過

#include<iostream>#include<string>using namespace std;string k;int x,n;void fun(string& a){	if(a=="0"){		k="0";		return;	}	int flag1=0;	int leng=a.size();	for(int i=0;i<leng;++i)	 if(a[i]!='0'&&a[i]!='.') flag1=1;	if(!flag1){		k="0";		return;	}	int t=0;	while(t<a.size()-1&&a[t]=='0'&&a[t+1]!='.') a.erase(a.begin());	int len=a.size();	int flag=1,pos=len;	for(int i=0;i<len;++i)	 if(a[i]=='.'){	 	pos=i;	 	flag=0;	 	break;	 }	if(flag) a+='.';	int y=100;	while(y--) a+='0';	len=a.size(); 	if(a[0]=='0'&&pos==1){		for(int i=2;i<len;++i)		 if(a[i]=='0') x--;		 else{		 	pos=i;		 	break;		 }		for(int i=pos;i<pos+n;++i) k+=a[i];	}	else{		x=pos;		int temp=0;		for(int i=0;i<len;++i){			if(a[i]!='.'){				k+=a[i];				temp++;				if(temp==n) break;			}		}	}}int main(){	string a,b,w,t;	int f,s;	cin>>n>>a>>b;	fun(a);	w=k;f=x;	k.clear();	x=0;	fun(b);	t=k;s=x;	if(w==t&&f==s){		cout<<"YES"<<' ';		if(w=="0"){			cout<<"0.";			for(int i=0;i<n;++i) cout<<"0";			cout<<"*10^"<<f;		}		else{			cout<<"0.";			cout<<w;			cout<<"*10^"<<f;		}	}	else{		cout<<"NO"<<' ';		if(w=="0"){			cout<<"0.";			for(int i=0;i<n;++i) cout<<"0";			cout<<"*10^"<<f;		}		else{			cout<<"0.";			cout<<w;			cout<<"*10^"<<f;		}		cout<<' ';		if(t=="0"){			cout<<"0.";			for(int i=0;i<n;++i) cout<<"0";			cout<<"*10^"<<f;		}		else{			cout<<"0.";			cout<<t;			cout<<"*10^"<<s;		}	}	return 0;}
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 12358.9Sample Output 1:
YES 0.123*10^5Sample Input 2:
3 120 128Sample Output 2:
NO 0.120*10^3 0.128*10^3