Educational Codeforces Round 5 A. Comparing Two Long Integers

A. Comparing Two Long Integers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two very long integers a, b (leading zeroes are allowed). You should check what number a or b is greater or determine that they are equal.

The input size is very large so don't use the reading of symbols one by one. Instead of that use the reading of a whole line or token.

As input/output can reach huge size it is recommended to use fast input/output methods: for example, prefer to use scanf/printfinstead of cin/cout in C++, prefer to use BufferedReader/PrintWriter instead of Scanner/System.out in Java. Don't use the function input() in Python2 instead of it use the function raw_input().

Input

The first line contains a non-negative integer a.

The second line contains a non-negative integer b.

The numbers a, b may contain leading zeroes. Each of them contains no more than 106 digits.

Output

Print the symbol "<" if a < b and the symbol ">" if a > b. If the numbers are equal print the symbol "=".

Examples

input

9
10

output

<

input

11
10

output

>

input

00012345
12345

output

=

input

0123
9

output

>

input

0123
111

output

>

字符串，去掉前面的零，比较大小，水题。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <string>
 using namespace std;
 typedef long long ll;
 typedef unsigned long long ull;
 typedef long double ld;

 int main()
 {
     char a[];
     char b[];
     cin >> a;
     cin >> b;
     int lena,lenb;
     int i,j;
     for(i = ; a[i] == ''; i++);
     for(j = ; b[j] == ''; j++);
     lena = strlen(a)-i;
     lenb = strlen(b)-j;
     if(lena ==  && lenb == )
     {
         cout << '=';
         return ;
     }
     else if(lena > lenb)
         cout << '>';
     else if(lena < lenb)
         cout << '<';
     else
     {
         int count = ;
         int k = strlen(a);
         for(int m = i, n = j;m < k; m++, n++)
         {
             int p = a[m] - '';
             int q = b[n] - '';
             if(p == q)
             {
                 count = ;
                 continue;
             }
             else if(p > q)
             {
                 count = ;
                 cout << '>';
                 break;
             }
             else
             {
                 count = ;
                 cout << '<';
                 break;
             }
         }
         if(count)
         {
             cout << '=';
         }
     }

     return ;
 }