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：
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,'\0',sizeof ra);
39 memset(apoint,'\0',sizeof apoint);
40 memset(rb,'\0',sizeof rb);
41 memset(bpoint,'\0',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)=='\0')&&strcmp(apoint,bpoint)==0)
47 puts("YES");
48 else if(*ra=='-'&&*(ra+1)=='\0'&&strcmp(apoint,bpoint)==0)
49 puts("YES");
50 else
51 puts("NO");
52 }
53 return 0;
54 }
View Code
编程是一种快乐，享受代码带给我的乐趣！！！
