功防世界reverse高手1
通过反汇编可得，flag需要对str_one和str_two进行xor运算（同0异1）
01001011011001001100101000010010101011001110011101010101010100010110110000010111100011110111001011010000010111010111000001100001
11101100110101000100011001000110110011111110010110011001010011101011111010110011010110111111100100100010111000100101110110111010
10100111101100001000110001010100011000110000001011001100000111111101001010100100110101001000101111110010101111110010110111011011经过转化得
a7b08c546302cc1fd2a4d48bf2bf2ddb（flag）