我需要生成4个随机数，每个数字在[-45 +45]度之间。如果rand％2 = 0，那么我想要结果（生成的随机数等于-angle）。一旦生成4个随机数，就需要扫描这些角度并找到锁定（角度相交的点）。 if语句中的循环中的-3，-2，-1，... + 3表示锁定发生在6度波束宽度内。代码有效。但它可以简化吗？另外，目标是通过扫描高度和两个点的方位角在两点之间建立锁定。
#include <iostream>
#include <conio.h>
#include <time.h>
using namespace std;
class Cscan
{
public:
int gran, lockaz, lockel;
};
int main()
{
srand (time(NULL));
int az1, az2, el1, el2, j, k;
BS1.lockaz = rand() % 46;
BS1.lockel = rand() % 46;
BS2.lockaz = rand() % 46;
BS2.lockel = rand() % 46;
k = rand() % 2;
if(k == 0)
k = -1;
BS1.lockaz = k*BS1.lockaz;
k = rand() % 2;
if(k == 0)
k = -1;
BS1.lockel = k*BS1.lockel;
k = rand() % 2;
if(k == 0)
k = -1;
BS2.lockaz = k*BS2.lockaz;
k = rand() % 2;
if(k == 0)
k = -1;
BS2.lockel = k*BS2.lockel;
for(az1=-45; az1<=45; az1=az1+4)
{
for(el1=-45; el1<=45; el1=el1+4)
{
for(az2=-45; az2<=45; az2=az2+4)
{
for(el2=-45; el2<=45; el2=el2+4)
{
if((az1==BS1.lockaz-3||az1==BS1.lockaz-2||az1==BS1.lockaz-1||az1==BS1.lockaz||az1==BS1.lockaz+1||az1==BS1.lockaz+2||az1==BS1.lockaz+3)&&
(az2==BS2.lockaz-3||az2==BS2.lockaz-2||az2==BS2.lockaz-1||az2==BS2.lockaz||az2==BS2.lockaz+1||az2==BS2.lockaz+2||az2==BS2.lockaz+3)&&
(el1==BS1.lockel-3||el1==BS1.lockel-2||el1==BS1.lockel-1||el1==BS1.lockel||el1==BS1.lockel+1||el1==BS1.lockel+2||el1==BS1.lockel+3)&&
(el2==BS2.lockel-3||el2==BS2.lockel-2||el2==BS2.lockel-1||el2==BS2.lockel||el2==BS2.lockel+1||el2==BS2.lockel+2||el2==BS2.lockel+3))
{
cout << "locked \n" << BS1.lockaz << " " << BS1.lockel << " " << BS2.lockaz << " " << BS2.lockel <<endl
< az1 << " " << el1 << " " << az2 << " " << el2 << endl;
k = 1;
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
_getch();
}
答案 0 :(得分：2)
BS1.lockaz = rand() % 91 - 45;
BS1.lockel = rand() % 91 - 45;
BS2.lockaz = rand() % 91 - 45;
BS2.lockel = rand() % 91 - 45;
答案 1 :(得分：1)
以度为单位的整数角度？非常值得怀疑。像物体一样的“物理”通常最好用浮点数表示，所以我先改变
typedef double angle;
struct Cscan { // why class? This is clearly POD
int gran; //I don't know what gran is. Perhaps this should also be floating-point.
angle lockaz, lockel;
};
这似乎使初看起来更加困难，因为%的随机范围选择不再起作用，也没有用于比较浮点数是否相等。 然而，这是一件好事，因为所有这些实际上都是非常糟糕的做法。
如果你想继续使用rand()作为随机数生成器（虽然我建议std::uniform_real_distribution），写一个函数来做到这一点：
const double pi = 3.141592653589793; // Let's use radians internally, not degrees.
const angle rightangle = pi/2.; // It's much handier for real calculations.
inline angle deg2rad(angle dg) {return dg * rightangle / 90.;}
angle random_in_sym_rightangle() {
return rightangle * ( ((double) rand()) / ((double) RAND_MAX) - .5 );
}
现在你要做
BS1.lockaz = random_in_sym_rightangle();
BS1.lockel = random_in_sym_rightangle();
BS2.lockaz = random_in_sym_rightangle();
BS2.lockel = random_in_sym_rightangle();
然后你需要进行范围检查。这又是一个专门的功能
bool equal_in_margin(angle theta, angle phi, angle margin) {
return (theta > phi-margin && theta < phi+margin);
}
然后你会对锁进行详尽的搜索。这肯定可以更有效地完成，但这是一个算法问题，与语言无关。坚持for循环，你仍然可以通过避免这种明确的中断检查使它们看起来更好。一种方法是好的goto，我建议你在这里加上一个额外的功能，并在你完成后返回
#define TRAVERSE_SYM_RIGHTANGLE(phi) \
for ( angle phi = -pi/4.; phi < pi/4.; phi += deg2rad(4) )
int lock_k // better give this a more descriptive name
( const Cscan& BS1, const Cscan& BS2, int k ) {
TRAVERSE_SYM_RIGHTANGLE(az1) {
TRAVERSE_SYM_RIGHTANGLE(el1) {
TRAVERSE_SYM_RIGHTANGLE(az2) {
TRAVERSE_SYM_RIGHTANGLE(el2) {
if( equal_in_margin( az1, BS1.lockaz, deg2rad(6.) )
&& equal_in_margin( el1, BS1.lockel, deg2rad(6.) )
&& equal_in_margin( az2, BS1.lockaz, deg2rad(6.) )
&& equal_in_margin( el2, BS2.lockel, deg2rad(6.) ) ) {
std::cout << "locked \n" << BS1.lockaz << " " << BS1.lockel << " " << BS2.lockaz << " " << BS2.lockel << '\n'
<< az1 << " " << el1 << " " << az2 << " " << el2 << std::endl;
return 1;
}
}
}
}
}
return k;
}