HDUOJ-----A == B ?

A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 49403    Accepted Submission(s): 7593

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

 

Author

8600 && xhd

 

Source

校庆杯Warm Up

 

此题解法：分整数和小数两部分来比较即可！！但要注意的事情还是比较多的

需要考虑：

+0.000    0.00

YES

+0.00 -0.00

YES

+0001.00 1

YES

+000.0000100  .00001

YES

代码如下：

 #include<cstdio>
 #include<cstring>
 #define MAX 20000
 char a[MAX],b[MAX];
 char ra[MAX],apoint[MAX],
      rb[MAX],bpoint[MAX];
 void func(char *a,char *ra,char *apoint)
 {
     int i=,k=;
     bool flag=true;
     int len=strlen(a);
     if(*a=='+')i++;
     else if(a[]=='-')
     {
         ra[k++]='-' ;
         i=k;
     }
     for( ; a[i]==''&&i<len ; i++ );

     for( ;a[i]!='.'&&i<len ; i++ )
     {
         ra[k++] = a[i] ;
     }
     int j;
          a[i]=='.'? i++ : i ;
    for(j=len-;a[j]==''&&j>=i;j--);

    for(k= ; i<=j ; i++)
     {
         apoint[k++]=a[i];
     }
 }

 int main( void )
 {
     while(scanf("%s%s",a,b)!=EOF)
     {
         memset(ra,'\0',sizeof ra);
         memset(apoint,'\0',sizeof apoint);
         memset(rb,'\0',sizeof rb);
         memset(bpoint,'\0',sizeof bpoint);
         func(a,ra,apoint);
         func(b,rb,bpoint);
       if(strcmp(ra,rb)==&&strcmp(apoint,bpoint)==)
           puts("YES");
       else if((*rb=='-'&&*(rb+)=='\0')&&strcmp(apoint,bpoint)==)
           puts("YES");
      else if(*ra=='-'&&*(ra+)=='\0'&&strcmp(apoint,bpoint)==)
                puts("YES");
      else
           puts("NO");
     }
     return ;
 }