1060 Are They Equal （25 分）

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 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

分析：

此題考查科學計數法。 1.確定指數。首先確定小數點的位置u和第一個非零元素的位置w（因為當數為小於0的小數時可能存在大量的前導0），當w等於字串的長度說明字串對應的浮點數是0（或者0.0000…00，即多個0），雖然對於0來說指數可以用任何數表示，但是當用科學計數法進行比較時會導致不同的0是不相等的，為避免此情況，將不同的0的指數統一設定為0，測試點6考查的是這個。 2.確定乘號前面的數。若存在前導0冊去除前導0，若數字不夠n個則尾部用0補足。 注：測試點往往是程式的邊界條件，本題第一次提交時，測試點6(index from 0)沒過，實則是在while迴圈結束後未處理若沒找到非零元素該如何如何的情況，漏掉了邊界條件。在程式設計時若遇到控制流的走向判斷（e.g. if條件，for,while迴圈，）條件要完備，處理走向要完備~~~

#include<iostream>
using namespace std;
int n;
void f(string &s,int &u){
	u=0;
	int len=s.length();
	while(u<len&&s[u]!='.') u++;
	if(u==len){
		int t=u;
		if(s=="0") u=0;
		for(int i=0;i<n-t;i++){
			s.push_back('0');
		}
	}else{ 
		int w=0;
		while(w<len&&(s[w]=='0'||s[w]=='.')) w++;
		if(w==len) w=u;
		s.erase(u,1);
		while(s[0]=='0') s.erase(0,1);
		len=s.length();
		for(int i=0;i<n-len;i++){
			s.push_back('0');
		}
		if(u<w) u=u-w+1;
		else u=u-w;
	}
	s="0."+s.substr(0,n);
}
int main(){
	string s1,s2;
	cin>>n>>s1>>s2;
	int u1=0,u2=0;
	f(s1,u1);
	f(s2,u2);
	if(u1==u2 && s1==s2){
		cout<<"YES "<<s1<<"*10^"<<u1;
	}else{
		cout<<"NO "<<s1<<"*10^"<<u1<<" "<<s2<<"*10^"<<u2;
	}
	return 0;
}