HDU 2054 A == B ?

A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 91260    Accepted Submission(s): 14572

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

一开始以为是很水的题来，直接用的double，结果没过，然后仔细一看，他没有位数限制，就该用的大数，

考虑到了前置0和后置0.

#include <stdio.h>
#include <string.h>

int main()
{
	int k1,k2,len1,len2,i,j,sign;
	char a[100055],b[100055];
	
	while(~scanf("%s%s",a,b))
	{
		len1=strlen(a);
		len2=strlen(b);
		k1=j=sign=0;
		for(i=0;i<len1;i++)
		{
			if(sign==0&&a[i]!='0')
				sign=1;
			if(sign==1)
				a[j++]=a[i];
			if(a[i]=='.')
				k1=1;
		}
		a[j]='\0';
		k2=j=sign=0;
		for(i=0;i<len2;i++)
		{
			if(sign==0&&b[i]!='0')
				sign=1;
			if(sign==1)
				b[j++]=b[i];
			if(b[i]=='.')
				k2=1;
		}
		b[j]='\0';
		len1=strlen(a);
		len2=strlen(b);
		for(i=len1-1;k1==1&&a[i]!='.';i--)
		{
			if(a[i]=='0'||a[i]=='.')
				a[i]='\0';
			else
				k1=0;
		}
		if(k1==1)
			a[i]='\0';
		for(i=len2-1;k2==1&&b[i]!='.';i--)
		{
			if(b[i]=='0'||b[i]=='.')
				b[i]='\0';
			else
				k2=0;
		}
		if(k2==1)
			b[i]='\0';
		if(strcmp(a,b)==0)
			printf("YES\n");
		else
			printf("NO\n"); 
	}
	
	return 0;
}