【Writeup】i春秋网络安全领域专项技能赛_Reverse_srcleak

【Writeup】i春秋网络安全领域专项技能赛_Reverse_srcleak

​ 本题给了一个cpp文件，其内容如下：

#include
using namespace std;

typedef unsigned int uint;



template  class IfElse;

template 
class IfElse {
public:
	using ResultType = MaybeA;
};

template 
class IfElse {
public:
	using ResultType = MaybeB;
};

template  struct func1 {
	enum { mid = (L + R + 1) / 2 };

	using ResultType = typename IfElse<(N < mid * mid),
		func1, func1 >::ResultType;

	enum { result = ResultType::result };
};

template  struct func1 { enum { result = L }; };

template  struct _func1 { enum { result = func1::result }; };









template
constexpr size_t func2 = (Input % 2) + func2< (Input / 2) >;

template<>
constexpr size_t func2<0> = 0;









template
constexpr size_t func3 = num % 2;









templatestruct NEXTN {
	const static uint value = ((n % m != 0) * n);
};
templatestruct NEXTM {
	const static uint value = (m * m <= n ? (m + 1) : 0);
};
templatestruct TEST {
	const static uint value = TEST::value, NEXTM::value>::value;
};
templatestruct TEST<0, m> {
	const static uint value = 0;
};
templatestruct TEST {
	const static uint value = 1;
};
templatestruct func4 {
	const static uint value = TEST::value;
};
template<>struct func4<1> {
	const static uint value = 0;
};
template<>struct func4<2> {
	const static uint value = 1;
};











int main(int argc, char**argv) {
	//input 5 uint numbers ,x1,x2,x3,x4,x5
	//the sum of them should be MIN


	cout << func3< func2 > << endl;
	cout << func3< func2 > << endl;
	cout << func3< func2 > << endl;
	cout << func3< func2 > << endl;
	cout << func3< func2 > << endl;

	// output: 1 1 1 1 1


	cout << _func1::result << endl;
	cout << _func1::result << endl;
	cout << _func1::result << endl;
	cout << _func1::result << endl;
	cout << _func1::result << endl;

	//output: 963 4396 6666 1999 3141




	//how many "1" will func4<1>,func4<2>,fun4<3>......fun4<10000> ::value  return?
	x6 = count;


	// your flag is flag{x1-x2-x3-x4-x5-x6}
	// if x1=1,x2=2,x3=3,x4=4,x5=5,x6=6
	// flag is     flag{1-2-3-4-5-6}



	return 0;
}

要求比较明确：读懂源码按条件计算得到x1-x6并以-连接即为flag。

笨一点的方法就是改写一下源码写出python脚本爆破出x1-x5，再计算出x6：

x1-x5：

def func2(x):
	if x == 0:
		return 0
	return (x % 2) + func2(x // 2)

def func3(x):
	return x % 2

def func1(N, L, R):
	if L == R:
		return L
	mid = (L + R + 1) // 2
	if N < mid * mid:
		return func1(N, L, mid - 1)
	else:
		return func1(N, mid, R)

def _func1(x):
	return func1(x, 1, x)



if __name__ == '__main__':
	x1_flag = False
	x2_flag = False
	x3_flag = False
	x4_flag = False
	x5_flag = False
	for i in range(10000000, 100000000):
		if func3(func2(i)) != 1:
			continue
		if _func1(i) == 963 and not x1_flag:
			print("x1:",i)
			x1_flag = True
		if _func1(i) == 4396 and not x2_flag:
			print("x2:",i)
			x2_flag = True
		if _func1(i) == 6666 and not x3_flag:
			print("x3:",i)
			x3_flag = True
		if _func1(i) == 1999 and not x4_flag:
			print("x4:",i)
			x4_flag = True
		if _func1(i) == 3141 and not x5_flag:
			print("x5:",i)
			x5_flag = True

x6：

def nextm(n, m):
	if m*m <= n:
		return m+1
	else:
		return 0

def nextn(n, m):
	return (n % m != 0) * n

def test(n, m):
	if n == 0:
		return 0
	if m == 0:
		return 1
	return test(nextn(n, m), nextm(n, m))

def func4(x):
	if x == 1:
		return 0
	if x == 2:
		return 1
	return test(x, 2)


if __name__ == '__main__':
	x6 = 0
	for i in range(1, 5):
		if func4(i*2-1) == 1:
			x6 += 1
	print(x6)

  快一点的话就是由语义和简单的测试可以得出，_func1的输入值是输出值的平方，即x1-x5可以由output的5个数分别平方获取；而x6是1-10000整数中质数的个数。

​ 最终flag：
​ flag{927369-19324816-44435556-3996001-9865881-1229}