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;
}