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
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：
1 #include<cstdio> 2 #include<cstring> 3 #define MAX 20000 4 char a[MAX],b[MAX]; 5 char ra[MAX],apoint[MAX], 6 rb[MAX],bpoint[MAX]; 7 void func(char *a,char *ra,char *apoint) 8 { 9 int i=0,k=0; 10 bool flag=true; 11 int len=strlen(a); 12 if(*a=='+')i++; 13 else if(a[0]=='-') 14 { 15 ra[k++]='-' ; 16 i=k; 17 } 18 for( ; a[i]=='0'&&i<len ; i++ ); 19 20 for( ;a[i]!='.'&&i<len ; i++ ) 21 { 22 ra[k++] = a[i] ; 23 } 24 int j; 25 a[i]=='.'? i++ : i ; 26 for(j=len-1;a[j]=='0'&&j>=i;j--); 27 28 for(k=0 ; i<=j ; i++) 29 { 30 apoint[k++]=a[i]; 31 } 32 } 33 34 int main( void ) 35 { 36 while(scanf("%s%s",a,b)!=EOF) 37 { 38 memset(ra,' ',sizeof ra); 39 memset(apoint,' ',sizeof apoint); 40 memset(rb,' ',sizeof rb); 41 memset(bpoint,' ',sizeof bpoint); 42 func(a,ra,apoint); 43 func(b,rb,bpoint); 44 if(strcmp(ra,rb)==0&&strcmp(apoint,bpoint)==0) 45 puts("YES"); 46 else if((*rb=='-'&&*(rb+1)==' ')&&strcmp(apoint,bpoint)==0) 47 puts("YES"); 48 else if(*ra=='-'&&*(ra+1)==' '&&strcmp(apoint,bpoint)==0) 49 puts("YES"); 50 else 51 puts("NO"); 52 } 53 return 0; 54 }