[英]rand () function C++
我需要生成4個隨機數,每個隨機數在[-45 +45]度之間。 如果rand%2 = 0,那么我想要結果(生成的隨機數等於-angle)。 一旦生成了4個隨機數,就需要掃描這些角度並找到一個鎖(角度相交的點)。 另外,if語句中的循環中的-3,-2,-1,... + 3表示鎖定發生在6度射束寬度內。 該代碼有效。 但是可以簡化嗎? 同樣的目的是在兩個點之間通過掃描高程和方位角在兩個點之間建立鎖定。
#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();
}
BS1.lockaz = rand() % 91 - 45;
BS1.lockel = rand() % 91 - 45;
BS2.lockaz = rand() % 91 - 45;
BS2.lockel = rand() % 91 - 45;
整數角度(度)? 非常值得懷疑。 通常將“物理”之類的角度最好表示為浮點數,因此我首先要進行更改
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;
};
乍一看,這似乎使它變得更加困難,因為帶有%
的隨機范圍選擇已不再起作用,而且比較浮點數是否相等也沒有太大用處。 但是,這是一件好事 ,因為實際上所有這些都是非常不好的做法。
如果您想繼續使用rand()
作為隨機數生成器(盡管我建議使用std::uniform_real_distribution
),請編寫一個函數來執行此操作:
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 );
}
現在你要做
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,您需要執行此范圍檢查。 這又是要放入專用功能的東西
bool equal_in_margin(angle theta, angle phi, angle margin) {
return (theta > phi-margin && theta < phi+margin);
}
然后,對鎖進行詳盡的搜索。 絕對可以更有效地完成此操作,但這是算法問題,與語言無關。 堅持使用for
循環,您仍然可以避免這種顯式的中斷檢查,使它們看起來更好。 一種方法是舊的goto
,我建議在這里將其粘貼在一個附加函數中,並在完成后返回
#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;
}
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