1060 Are They Equal （25 point(s)）

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

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

经验总结：

这一题，难度还可以，就是细节的处理让人摸不着头脑，题目也没有任何提示，而且题目的测试集还不严谨.....很想跟他们反馈却根本找不到反馈的地方（遇到不严谨的地方不止一次了）
首先，说说这一题的不严谨的地方吧，对于最低限度不严谨的AC的代码，如果你输入  3 123 0.123   ，竟然可以得出   这两个浮点数是相等的结论.....  很明显，是没有判断指数是否相等。
然后，就是要注意这样一个样例 3 000002.0000001 2 这两个浮点数是相等的，就是注意前导0的处理。
最后，要注意0的处理，比如输出样例 3 0 000000.0000000，应该相等，输出的时候还要按照指定形式进行输出，输出应该为，YES 0.000*10*0
就这么多啦~

AC代码

#include <cstdio>
#include <cstring>
const int maxn=110;

int find(char str[])
{
  int i;
  for(i=0;str[i]!='\0';++i)
    if(str[i]=='.')
      return i;
  return i;
}
int dispose(char a[],int n,char c[],int &pos)
{
  int num=0;
  int len=strlen(a);
  int xpos=0;
  while(a[xpos]=='0'&&xpos<pos)
    ++xpos;
  for(int i=xpos;i<pos&&num<n;++i)
    c[num++]=a[i];
  if(n-num>0)
  {
    if(num==0)
    {
      xpos=pos+1;
      while(a[xpos]=='0')
        ++xpos;
      for(int i=xpos;i<len&&num<n;++i)
        c[num++]=a[i];
    }
    else
      for(int i=pos+1;i<len&&num<n;++i)
        c[num++]=a[i];
  }
  if(num==0)
  {
    pos=0;
    xpos=0;
  }
  int x=n-num;
  for(int i=0;i<x;++i)
    c[num++]='0';
  c[num]='\0';
  return xpos;
}
int main()
{
  int n,apos0,bpos0;
  char a[maxn],b[maxn],coma[maxn],comb[maxn];
  scanf("%d %s %s",&n,a,b);
  int posa=find(a);
  int posb=find(b);
  apos0=dispose(a,n,coma,posa);
  bpos0=dispose(b,n,comb,posb);
  if(posa-apos0<0)
    ++posa;
  if(posb-bpos0<0)
    ++posb;
  if(strcmp(coma,comb)==0&&posa-apos0==posb-bpos0)
    printf("YES 0.%s*10^%d\n",coma,posa-apos0);
  else
    printf("NO 0.%s*10^%d 0.%s*10^%d\n",coma,posa-apos0,comb,posb-bpos0);
  return 0;
}