HDU2054：A == B ?

A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 60827    Accepted Submission(s): 9451

Problem Description

Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO".

Input

each test case contains two numbers A and B.

Output

for each case, if A is equal to B, you should print "YES", or print "NO".

Sample Input

1 2
2 2
3 3
4 3

Sample Output

NO
YES
YES
NO

解题思路：

1. 一看题目就觉得恶心，题目条件越简单，ＡＣ的条件就越苛刻。被坑了一次

2.关键在于小数点，去除前导0和多余的后置0。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2054

 #include<stdio.h>
 #include<string.h>
 char n[], m[];
 int main()
 {
     int i, j, len_n, len_m, flag, len_last, len_late,doc, q, p, count, start;
     char ch;
     while()
     {
         memset(n, , sizeof(n));
         memset(m, , sizeof(m));
         flag = i = j = doc = count = start = ;
         while()
         {
             if(scanf("%c", &ch) == EOF) return ;
             if(ch == '\n' && start == ) break;
             else if(ch != ' ' && ch != '\n' && flag == )
             {
                 n[i++] = ch;
             }
             else if(ch == ' ' || ch == '\n')
             {
                 flag = ;
                 start = ;
             }
             else m[j++] = ch;

         }
         len_n = strlen(n);
         len_m = strlen(m);
         if((n[] == '-' && m[] != '-') || (n[] != '-' && m[] == '-'))
         {
             printf("NO\n");
             continue;
         }
         for(i = n[] != '-'? :; i<len_n && n[i] - ''== ; i++);
         for(j = m[] != '-'? :; j<len_m && m[j] - ''== ; j++);
         len_n -= i;
         len_m -= j;
         len_last = len_n < len_m? len_n:len_m;
         len_late = len_n + len_m - len_last;
         {
             for(; count < len_last; i++, j++, count++)
             if(n[i] - '' != m[j] - '') break;
             else if(n[i] == '.') doc = ;
             if(count >= len_late)
             {

                 printf("YES\n");
             }
             else
             {
              if(len_n > len_m)
              {
                  if(doc !=  && n[i] != '.') printf("NO\n");
                  else
                  {
                      for(i = n[i] == '.'? i+: i; i<strlen(n); i++)
                      if(n[i] - '' !=  ) break;
                      if(i>=len_n) printf("YES\n");
                      else printf("NO\n");
                  }

              }
              else if(len_n < len_m)
              {
                 if(doc !=  && m[j] != '.') printf("NO\n");
                  else
                  {
                      for(j = m[j] == '.'? j+: j; j<strlen(m); j++)
                      if(m[j] - '' !=  ) break;
                      if(j>=len_m) printf("YES\n");
                      else printf("NO\n");
                  }
              }
              else if(len_n == len_m) printf("NO\n");
             }
         }

     }
     return ;
 }