A == B ?

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 133531 Accepted Submission(s): 21293

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

这题真是一个巨坑

因为题中没有给出A，B是什么样的数，所以需要考虑的不仅仅是 大数 的问题还要考虑 小数 的问题。

我一开始没注意到小数点后还有数要去比就直接把小数点换成'\0'结果就WA了好几次（2333）；

代表测试样例

0.0 0
YES
1.222 1
NO

样例代码

#include <bits/stdc++.h>
using namespace std;
char a[100000],b[100000];
int main()
{
    while(~scanf("%s%s",a,b))
    {
        int oja=0,ojb=0;
        int lena=strlen(a);
        int lenb=strlen(b);
        for(int i=0; i <= lena-1; i++)
            if(a[i] == '.')
                oja=1;
        for(int i=0; i <= lenb-1; i++)
            if(b[i] == '.')
                ojb=1;
        if(oja == 1)        //下面的可以单独定义一个函数，不过TL不TL就不知道了
        {
            while(a[lena-1] == '0')
            {
                a[lena-1]='\0';
                lena--;
            }
            if(a[lena-1] == '.')
                a[lena-1] = '\0';
        }
        if(ojb == 1)
        {
            while(b[lenb-1] == '0')
            {
                b[lenb-1]='\0';
                lenb--;
            }
            if(b[lenb-1] == '.')
                b[lenb-1] = '\0';
        }
        if(strcmp(a,b) == 0)
            cout << "YES" << endl;
        else
            cout << "NO" << endl;
    }
    return 0;
}