PAT 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 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.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

结尾无空行

Sample Output 1:

YES 0.123*10^5

结尾无空行

Sample Input 2:

3 120 128

结尾无空行

Sample Output 2:

NO 0.120*10^3 0.128*10^3

结尾无空行

思路：输入数据是有前导0的，首先去除前导0化成xxx.xx...x的形式；再考虑有没有小数点，
有小数点的话再看小数点是否在第二位，即x.xx...x的形式，然后通过移位对指数
进行加减，最后去掉小数点（最后的结果都为0.xx..x,相当于全不加0.）。比较的时候
将指数和数值部分一起比较。

本题坑点主要有三个：前导0；0和0.0；最后数值位长度恒为n（不足要补0，多了要截断）

附上一些测试点：

4 0000 0000.00        YES 0.0000*10^0
4 00123.5678 0001235        NO 0.1235*10^3 0.1235*10^4
3 0.0520 0.0521        NO 0.520*10^-1 0.521*10^-1
4 00000.000000123 0.0000001230        YES 0.1230*10^-6
4 00100.00000012 100.00000013        YES 0.1000*10^3
5 0010.013 10.012        NO 0.10013*10^2 0.10012*10^2
4 123.5678 123.5        YES 0.1235*10^3
3 123.5678 123        YES 0.123*10^3
4 123.0678 123        YES 0.1230*10^3
3 0.000 0        YES 0.000*10^0

代码如下：

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,exa=0,exb=0,posa=-1,posb=-1;
    string a,b;
    scanf("%d ",&n);
    cin>>a>>b;
    posa=a.find('.');posb=b.find('.');
    if(posa!=-1)
    {
        while(a.length()>0&&a[0]=='0'&&a[1]!='.') a.erase(a.begin());
        posa=a.find('.');
        if(posa==1)
        {
            if(a[0]=='0')
            {
                int temp=posa+1;
                while(temp<a.length()&&a[temp]=='0')
                {
                    exa--;
                    temp++;
                }
                a.erase(a.begin()+posa);
                while(a.length()>0&&a[0]=='0') a.erase(a.begin());
                if(a.length()==0) a="0";
            }
            else
            {
                exa++;
                a.erase(a.begin()+posa);
            }
        }
        else
        {
            exa+=posa;
            a.erase(a.begin()+posa);
        }
    }
    else
    {
        while(a.length()>0&&a[0]=='0') a.erase(a.begin());
        exa+=a.length();
        if(a.length()==0) a="0";
    }
    if(posb!=-1)
    {
        while(b.length()>0&&b[0]=='0'&&b[1]!='.') b.erase(b.begin());
        posb=b.find('.');
        if(posb==1)
        {
            if(b[0]=='0')
            {
                int temp=posb+1;
                while(temp<b.length()&&b[temp]=='0')
                {
                    exb--;
                    temp++;
                }
                b.erase(b.begin()+posb);
                while(b.length()>0&&b[0]=='0') b.erase(b.begin());
                if(b.length()==0) b="0";
            }
            else
            {
                exb++;
                b.erase(b.begin()+posb);
            }
        }
        else
        {
            exb+=posb;
            b.erase(b.begin()+posb);
        }
    }
    else
    {
        while(b.length()>0&&b[0]=='0') b.erase(b.begin());
        exb+=b.length();
        if(b.length()==0) b="0";
    }
    if(a=="0") exa=0;
    if(b=="0") exb=0;
    string s1=a.substr(0,n);
    string s2=b.substr(0,n);
    int len1=s1.length(),len2=s2.length();
    int len=max(max(len1,len2),n);
    while(s1.length()<len) s1+="0";
    while(s2.length()<len) s2+="0";
    if(exa==exb&&s1==s2) printf("YES 0.%s*10^%d",s1.c_str(),exa);
    else printf("NO 0.%s*10^%d 0.%s*10^%d",s1.c_str(),exa,s2.c_str(),exb);
    return 0;
}