Hiya,我的代码目前有三个函数,每个函数都产生一个大量的随机数。我想知道是否有只是有一个函数返回一个链表或多维数组使它有点整洁的办法:有没有使用链表来简化我的蒙特卡洛代码
(从http://pastebin.com/Y5aE6XKS复制)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#ifndef RAND_MAX
#define RAND_MAX 2147483648
#endif
#define N 420000
double* rdm_X(void);
double* rdm_Y(void);
double* rdm_Z(void);
void main(void)
{
double* Random_number_list_X = rdm_X();
double* Random_number_list_Y = rdm_Y();
double* Random_number_list_Z = rdm_Z();
double X[N+1], Y[N+1], Z[N+1], density = 1, vol = 42.0;
double sum = 0, sum_x = 0, sum_y = 0, sum_z = 0;
int i;
for (i = 0; i <= N; i++) {
X[i] = 3 * Random_number_list_X[i] + 1;
Y[i] = 7 * Random_number_list_Y[i] - 3;
Z[i] = 2 * Random_number_list_Z[i] - 1;
if ((Z[i]*Z[i]) + (sqrt(X[i]*X[i] + Y[i]*Y[i]) - 3)*(sqrt(X[i]*X[i] + Y[i]*Y[i]) - 3) <= 1) {
sum += density;
sum_x += X[i] * density;
sum_y += Y[i] * density;
sum_z += Z[i] * density;
}
}
printf("(%.5lf, %.5lf, %.5lf)\n",
sum_x/sum, sum_y/sum, sum_z/sum);
}
double* rdm_X(void)
{
double* Random_number_list_X = calloc(N + 1, sizeof(double));
int i;
srand(time(NULL));
for (i = 1; i <= N; i++) {
Random_number_list_X[i] = (float) rand()/(float) RAND_MAX;
}
return Random_number_list_X;
}
double* rdm_Y(void)
{
double* Random_number_list_Y = calloc(N + 1, sizeof(double));
int i;
sleep(1);
srand(time(NULL));
for (i = 1; i <= N; i++) {
Random_number_list_Y[i] = (float) rand()/(float) RAND_MAX;
}
return Random_number_list_Y;
}
double* rdm_Z(void)
{
double* Random_number_list_Z = calloc(N + 1, sizeof(double));
int i;
sleep(2);
srand(time(NULL));
for (i = 1; i <= N; i++) {
Random_number_list_Z[i] = (float) rand()/(float) RAND_MAX;
}
return Random_number_list_Z;
}
你问了六个问题,从未接受答案。试着去做点什么! – 2010-09-10 20:02:45
为什么你的代码是在pastebin而不是在这里的任何原因? – pmg 2010-09-10 20:08:33
我没有意识到我不得不,我现在去做,欢呼 – JMzance 2010-09-11 18:53:07