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    42号混凝土应该拌意大利面
    意大利面要拌42号混凝土,因为螺丝钉的长度影响挖掘机的扭矩,往下砸会产生高能蛋白ufo,非常影响经济发展!甚至对太平洋和充电器都会造成核污染!又根据勾股定理icon可以推断出,人工饲养的东条英鸡可以捕获野生三角函数!所以不管秦始皇的切面是否具有放射性,还是特朗普icon的n次方含有沉淀物!都不影响沃尔玛和维尔康在南极汇合!
    作品
    #include<bits/stdc++.h>
    using namespace std;
    void kst(int x,int y,int z,int s)
    {
    	char ys;
    	x = rand()% z;
    	y = rand()% z;
    	if(s == 1)
    	{
    		ys = '+';
    		cout << x << ys << y << "=" << endl;
    	}
    	else if(s == 21)
    	{
    		ys = '-';
    		if(x >= y)
    			cout << x << ys << y << "=" << endl;
    		else
    			cout << y << ys << x << "=" << endl;
    	}
    	else if(s == 22)
    	{
    		ys = '-';
    		if(x <= y)
    			cout << x << ys << y << "=" << endl;
    		else
    			cout << y << ys << z << "=" << endl;
    	}
    	else if(s == 23)
    	{
    		ys = '-';
    		cout << x << ys << y << "=" << endl;
    	}
    	else if(s == 3)
    	{
    		ys = '*';
    		cout << x << ys << y << "=" << endl;
    	}
    	else if(s == 41)
    	{
    		ys = '/';
    		if(x % y != 0 && x == 0 && y == 0)
    		{
    			while(x % y != 0 && x == 0 && y == 0)
    			{
    				x = rand()% z;
    				y = rand()% z;
    			}
    			cout << x << ys << y << "=" << endl;
    		}
    		else
    			cout << x << ys << y << "=" << endl;
    	}
    	else if(s == 42)
    	{
    		ys = '/';
    		if(x % y == 0 && x == 0 && y == 0)
    		{
    			while(x % y == 0 && x == 0 && y == 0)
    			{
    				x = rand()% z;
    				y = rand()% z;
    			}
    			cout << x << ys << y << "=" << endl;
    		}
    		else
    		{
    			cout << x << ys << y << "=" << endl;
    		}
    	}
    	else if(s == 43)
    	{
    		ys = '/';
    		cout << x << ys << y << "=" << endl;
    	}
    	else
    	{
    		s = rand() % 4;
    		if(s == 0)
    		{
    			cout << x << '+' << y << "=" << endl;
    		}
    		else if(s == 1)
    		{
    			cout << x << '-' << y << "=" << endl;
    		}
    		else if(s == 2)
    		{
    			cout << x << '*' << y << "=" << endl;
    		}
    		else
    		{
    			cout << x << '/' << y << "=" << endl;
    		}
    	}
    }
    int a,b,c,n,f;
    int main() {
    	cout << "请输入范围(<=32767)" << endl;
    	cin >> c;
    	cout << "请输入道数(<=2147483647)" << endl;
    	cin >> n;
    	cout << "请输入算法(1.+)(21.-|得数为正|)(22.-|得数为负|)(23.-|得数随机|)(3.*)(41./|得数为整数|)(42./|得数为小数[分数]|)(43./|得数随机|)(5.算法随机)" << endl;
    	cin >> f;
    	freopen("口算题.txt","w",stdout);
    	srand(time(0));
    	for(int i = 1;i <= n;i++) 
    	{
    		kst(a,b,c,f);
    	}
    	return 0;
    }
    
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