[ctf.show.reverse] 数学不及格

主要函数

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v4; // [rsp+14h] [rbp-4Ch]
  char *endptr; // [rsp+18h] [rbp-48h] BYREF
  char *v6; // [rsp+20h] [rbp-40h] BYREF
  char *v7; // [rsp+28h] [rbp-38h] BYREF
  char *v8; // [rsp+30h] [rbp-30h] BYREF
  __int64 v9; // [rsp+38h] [rbp-28h]
  __int64 v10; // [rsp+40h] [rbp-20h]
  __int64 v11; // [rsp+48h] [rbp-18h]
  __int64 v12; // [rsp+50h] [rbp-10h]
  unsigned __int64 v13; // [rsp+58h] [rbp-8h]

  v13 = __readfsqword(0x28u);
  if ( argc != 5 )
  {
    puts("argc nonono");
    exit(1);
  }
  v4 = strtol(argv[4], &endptr, 16) - 0x6543;
  v9 = f(v4);                                   // 斐波那契第n项
  v10 = strtol(argv[1], &v6, 16);
  v11 = strtol(argv[2], &v7, 16);
  v12 = strtol(argv[3], &v8, 16);
  if ( v9 - v10 != 0x233F0E151CLL )
  {
    puts("argv1 nonono!");
    exit(1);
  }
  if ( v9 - v11 != 0x1B45F81A32LL )
  {
    puts("argv2 nonono!");
    exit(1);
  }
  if ( v9 - v12 != 0x244C071725LL )
  {
    puts("argv3 nonono!");
    exit(1);
  }
  if ( v4 + v12 + v11 + v10 != 0x13A31412F8CLL )
  {
    puts("argv sum nonono!");
    exit(1);
  }
  puts("well done!decode your argv!");
  return 0;
}

实际上就是一个数学计算，翻译一下就是1到4，4个式子相加（小学数学问题）得到5，然后爆到58

fib(a4-0x6543) - a1      = 0x233F0E151C
fib(a4-0x6543) - a2      = 0x1B45F81A32
fib(a4-0x6543) - a3      = 0x244C071725
a4-0x6543 + a3 + a2 + a1 = 0x13A31412F8C
a4-0x6543 + 3* fib(a4-0x6543) = sum()

def fib(n):
    a=1
    b=1
    i=2
    while i