A == B ?（思维）

A == B ?

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 121015    Accepted Submission(s): 19354

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

思路：把小数点后面的多余的零去掉，同时注意如果最后一位是小数点的话，也要去掉，（就在这里wa了好多次）

上代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e6+5;
char a[maxn],b[maxn];
int qSpot(char *c,int lenn){
	int len = lenn,len1 = lenn;
	for(int i = 0;i<len;i++){
		if(c[i]=='.'){
			len = i;
			break;
		}
	}
	int s = 0;
	for(int i = len1-1;i>=len;i--){
		if(c[i]=='0'){
			s++;
		}
		else{
			break;
		}
	}
	if(c[len1-s-1] == '.') s++;
	return len1 - s;
}
int main()
{
	while(~scanf("%s %s",a,b)){
		int len = strlen(a);
		int len1 = strlen(b);
		len = qSpot(a,len);
		len1 = qSpot(b,len1);
		//printf("%d %d\n",len,len1);
		if(len != len1) printf("NO\n");
		else{
			bool falg = true;
			for(int i = 0;i<len;i++){
				if(a[i]!=b[i]){
					falg = false;
				}
				if(falg == false) break;
			}
			if(falg) printf("YES\n");
			else printf("NO\n");
		}
	}
	return 0;
 }