我正在做炸弹实验室，我必须通过调试反汇编的二进制文件来化解炸弹。但是我不知道如何调试此阶段以获得答案。帮助将不胜感激。如果有人可以找出答案，那将是很好，但是一些见识也将是很好。
第六阶段
0x08048e13 <+0>: sub $0x2c,%esp
0x08048e16 <+3>: lea 0x1c(%esp),%eax
0x08048e1a <+7>: mov %eax,0xc(%esp)
0x08048e1e <+11>: lea 0x18(%esp),%eax
0x08048e22 <+15>: mov %eax,0x8(%esp)
0x08048e26 <+19>: movl $0x804a5b1,0x4(%esp)
0x08048e2e <+27>: mov 0x30(%esp),%eax
0x08048e32 <+31>: mov %eax,(%esp)
0x08048e35 <+34>: call 0x8048870 <__isoc99_sscanf@plt>
0x08048e3a <+39>: cmp $0x2,%eax
0x08048e3d <+42>: jne 0x8048e4b <phase_6+56>
0x08048e3f <+44>: mov 0x1c(%esp),%eax
0x08048e43 <+48>: sub $0x2,%eax
0x08048e46 <+51>: cmp $0x2,%eax
0x08048e49 <+54>: jbe 0x8048e50 <phase_6+61>
0x08048e4b <+56>: call 0x8049395 <explode_bomb>
0x08048e50 <+61>: mov 0x1c(%esp),%eax
0x08048e54 <+65>: mov %eax,0x4(%esp)
0x08048e58 <+69>: movl $0x7,(%esp)
0x08048e5f <+76>: call 0x8048dc9 <func6>
0x08048e64 <+81>: cmp 0x18(%esp),%eax
0x08048e68 <+85>: je 0x8048e6f <phase_6+92>
0x08048e6a <+87>: call 0x8049395 <explode_bomb>
0x08048e6f <+92>: add $0x2c,%esp
0x08048e72 <+95>: ret
功能6
0x08048dc9 <+0>: push %edi
0x08048dca <+1>: push %esi
0x08048dcb <+2>: push %ebx
0x08048dcc <+3>: sub $0x10,%esp
0x08048dcf <+6>: mov 0x20(%esp),%ebx
0x08048dd3 <+10>: mov 0x24(%esp),%esi
0x08048dd7 <+14>: test %ebx,%ebx
0x08048dd9 <+16>: jle 0x8048e07 <func6+62>
0x08048ddb <+18>: mov %esi,%eax
0x08048ddd <+20>: cmp $0x1,%ebx
0x08048de0 <+23>: je 0x8048e0c <func6+67>
0x08048de2 <+25>: mov %esi,0x4(%esp)
0x08048de6 <+29>: lea -0x1(%ebx),%eax
0x08048de9 <+32>: mov %eax,(%esp)
0x08048dec <+35>: call 0x8048dc9 <func6>
0x08048df1 <+40>: lea (%eax,%esi,1),%edi
0x08048df4 <+43>: mov %esi,0x4(%esp)
0x08048df8 <+47>: sub $0x2,%ebx
0x08048dfb <+50>: mov %ebx,(%esp)
0x08048dfe <+53>: call 0x8048dc9 <func6>
0x08048e03 <+58>: add %edi,%eax
0x08048e05 <+60>: jmp 0x8048e0c <func6+67>
0x08048e07 <+62>: mov $0x0,%eax
0x08048e0c <+67>: add $0x10,%esp
0x08048e0f <+70>: pop %ebx
0x08048e10 <+71>: pop %esi
0x08048e11 <+72>: pop %edi
0x08048e12 <+73>: ret
End of assembler dump.