hdu 2054

A == B ?

[align=left]Problem Description[/align]Give you two numbers A and B, if A is equal to B, you should print "YES", or print "NO". 
[align=left]Input[/align]each test case contains two numbers A and B.
 
[align=left]Output[/align]for each case, if A is equal to B, you should print "YES", or print "NO". 
[align=left]Sample Input[/align]

1 22 23 34 3 [align=left]Sample Output[/align]

NOYESYESNO 

/*
分析 ： 输入两个数A 、B 判断两个数是否相等。
未给数据范围 所以我们应该考虑用字符串存储来处理它们
因为题目可能会出前导零 后导零的情况 而且最前面可能会有‘+’ ‘-’
例如 001 与 0001 这两个是相等的
001.00000 与 1 也是相等的
+001.000 与 1 相等
+1 与 -1 不相等
所以我先找出小数点的位置
temp=a.find('.');0~temp 与 temp~a.length();
分别扫一遍，进行处理字符串。
然后进行比较即可。
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <cstdlib>
using namespace std;
string a,b;
void work(string a,string b,int len1,int len2)
{
int temp1=a.find('.'),temp2=b.find('.');
int i,j,k;
int front_zero_1=0,back_zero_1=0;
if(temp1!=-1)//有逗号
{
for(i=0;i<temp1-1;i++)
{
if(a[i]=='+'||a[i]=='-') continue;
if(a[i]!='0') break;
if(a[i]=='0')
{
front_zero_1=1;
}
}
if(front_zero_1==1)
{
if(a[0]=='+'||a[0]=='-') a.erase(a.begin()+1,a.begin()+i);
else a.erase(a.begin(),a.begin()+i);
}
temp1=a.find('.');
len1=a.length();
for(i=len1-1;i>temp1;i--)
{
if(a[i]!='0') break;
if(a[i]=='0')
{
back_zero_1=1;
continue;
}
}
if(back_zero_1) a.erase(a.end()-len1+i+1,a.end());
}
else
{
for(i=0;i<len1;i++)
{
if(a[i]=='+'||a[i]=='-') continue;
if(a[i]!='0') break;
if(a[i]=='0')
{
front_zero_1=1;
continue;
}
}
if(front_zero_1)
{
if(a[0]=='+'||a[0]=='-') a.erase(a.begin()+1,a.begin()+i);
else a.erase(a.begin(),a.begin()+i);
}
}
int front_zero_2=0,back_zero_2=0;
if(temp2!=-1)//有逗号
{
for(i=0;i<temp2-1;i++)
{
if(b[i]=='+'||b[i]=='-') continue;
if(b[i]!='0') break;
if(b[i]=='0')
{
front_zero_2=1;
continue;
}
}
if(front_zero_2)
{
if(b[0]=='+'||b[0]=='-') b.erase(b.begin()+1,b.begin()+i);
else b.erase(b.begin(),b.begin()+i);
}
temp2=b.find('.');
len2=b.length();
for(i=len2-1;i>temp2;i--)
{
if(b[i]!='0') break;
if(b[i]=='0')
{
back_zero_2=1;
continue;
}
}
if(back_zero_2) b.erase(b.end()-len2+i+1,b.end());
}
else
{
for(i=0;i<len2;i++)
{
if(b[i]=='+'||b[i]=='-') continue;
if(b[i]!='0') break;
if(b[i]=='0')
{
front_zero_2=1;
continue;
}
}
if(front_zero_2)
{
if(b[0]=='+'||b[0]=='-') b.erase(b.begin()+1,b.begin()+i);
else b.erase(b.begin(),b.begin()+i);
}
}
if(a.find('.')+1==a.length()) a.erase(a.find('.'));
if(b.find('.')+1==b.length()) b.erase(b.find('.'));
if(a[0]=='+') a.erase(a.begin());
if(b[0]=='+') b.erase(b.begin());
//cout<<front_zero_1<<" "<<front_zero_2<<endl;
//cout<<back_zero_1<<" "<<back_zero_2<<endl;
//cout<<a<<endl;
//cout<<b<<endl;
if(a==b) puts("YES");
else puts("NO");
}
int main()
{
while(cin>>a>>b)
work(a,b,a.length(),b.length());
return 0;
}