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(shù)法。
1.首先確定小數(shù)點的位置u和第一個非零元素的位置w（因為當數(shù)為小于0的小數(shù)時可能存在大量的前導0），當w等于字符串的長度說明字符串對應的浮點數(shù)是0（或者0.0000..00，即多個0），雖然對于0來說指數(shù)可以用任何數(shù)表示，但是當用科學計數(shù)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(shù)統(tǒng)一設定為0，測試點6考查的是這個。
注：測試點往往是程序的邊界條件，本題第一次提交時，測試點6沒過，實則是在while循環(huán)結(jié)束后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。在編程時若遇到控制流的走向判斷（e.g. if條件，for,while循環(huán)，）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;
}