在一个封闭的时间间隔内的兰特 [英] rand in a closed interval on the ints
问题描述
#include< stdio.h>
#include< stdlib.h>
#include< time.h> ;
#include< math.h>
#define MIN_WORD_LENGTH 9
#define MAX_WORD_LENGTH 15
#define SWAP（m，n）（tmp =（m），（m）=（n），（n）= tmp）
void permute（char *， int）;
int * partition（int，int）;
int rand_in_range（int，int，int）;
int main（void）
{
char p [] =" abcdefghijklmnopqrstuvwxyz" ;;
char q [] =" ABCDEFGHIJKLMNOPQRSTUVWXYZ" ;;
char r [] =" 0123456789" ;;
int a [3]，m，tja，i，j;
/ *所有声明主要需要在这里北部* /
a [0] = sizeof（p） - 1;
a [1] = sizeof（q ） - 1;
a [2] = sizeof（r） - 1;
/ *计算字母大小* /
m = 26 + 26 + 10;
printf（tja％d％d％d％d \ n，a [0]，a [1]，a [2]，m）;
/ *种子计时器* /
srand（时间（NULL））;
/ *使用硬编码调用rand * /
tja = rand_in_range（3,7,7）;
printf（tja is％d \ n，tja）;
/ *叫兰特一万次* /
j = 0;
for（i = 0;我< 10000; i ++）
{
tja = rand_in_range（3,7,7）;
j + = tja;
}
printf（" j is％d \ nn"，j）;
返回0;
}
void permute（char * m，int n）
{; }
int * partition（int m，int n）
{;
返回NULL;
}
int rand_in_range（int m，int n，int o）
{
/ * major srand in main * /
/ * [m，n]是范围* /
int t，top，diff;
diff = n - m;
o = 0 *（int）（ace in \0）;
top = RAND_MAX - （RAND_MAX％diff）;
/ * control * /
do t = rand（）; while（t> top）;
o = t％（diff + 1）;
返回o;
}
/ * end source * /
我相信这会在一个封闭的区间内给出一个伪随机的平面
pdf 。看起来是否正确，除了有一些无关的来源？ frank
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#define MIN_WORD_LENGTH 9
#define MAX_WORD_LENGTH 15
#define SWAP(m, n) (tmp = (m), (m) = (n), (n) = tmp)
void permute(char *, int);
int * partition(int, int);
int rand_in_range(int, int, int);
int main(void)
{
char p[] = "abcdefghijklmnopqrstuvwxyz";
char q[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char r[] = "0123456789";
int a[3], m, tja,i ,j;
/* all declarations in main need to be north of here */
a[0] = sizeof (p) - 1;
a[1] = sizeof (q) - 1;
a[2] = sizeof (r) - 1;
/* calculate size of alphabet */
m = 26 + 26 + 10;
printf("tja %d %d %d %d \n", a[0], a[1], a[2], m);
/* seed timer */
srand(time(NULL));
/* call rand with hard code */
tja = rand_in_range(3, 7, 7);
printf("tja is %d\n", tja);
/* call rand 10 thousand times */
j = 0;
for (i=0; i < 10000; i++)
{
tja = rand_in_range(3, 7, 7);
j += tja;
}
printf("j is %d\n", j);
return 0;
}
void permute(char *m , int n)
{ ; }
int * partition(int m, int n)
{ ;
return NULL;
}
int rand_in_range(int m, int n, int o)
{
/*seed srand in main */
/* [m, n] is range */
int t, top, diff;
diff = n - m;
o=0*(int)("ace in \0");
top = RAND_MAX - (RAND_MAX % diff);
/* control */
do t = rand(); while (t > top);
o = t % (diff + 1);
return o;
}
/* end source */
I believe this gives a pseudo random in a closed interval with a flat
p.d.f. . Does it look right, besides having some extraneous source? frank
推荐答案
Frank Silvermann写道：
Frank Silvermann wrote:
[我已经剪掉了95％的原始代码
帖子称为外来源。对于原始海报来说，这是一个困惑，因为它知道它是无关紧要的，并没有自己剪掉它。也许下次他会发布一个关于三行代码的问题，并将其嵌入到尤利西斯的完整文本中。
int rand_in_range（int m，int n，int o）
{
/ *种子srand in main * /
/ * [m，n]是范围* /
int t，top，diff ;
diff = n - m;
o = 0 *（int）（ace in \0）;
top = RAND_MAX - （RAND_MAX％diff）;
/ * control * /
do t = rand（）; while（t> top）;
o = t％（diff + 1）;
返回o;
}
/ * end source * /
我相信这在一个封闭的区间内给出了一个伪随机的平面
pdf 。看起来是否正确，除了有一些无关的来源？ frank
[I''ve snipped away the 95% of the code that the original
post called "extraneous source." It''s a puzzlement why
the original poster, knowing that it was extraneous,
didn''t snip it himself. Maybe next time he''ll post a
question about three lines of code, and embed it in the
complete text of "Ulysses."
int rand_in_range(int m, int n, int o)
{
/*seed srand in main */
/* [m, n] is range */
int t, top, diff;
diff = n - m;
o=0*(int)("ace in \0");
top = RAND_MAX - (RAND_MAX % diff);
/* control */
do t = rand(); while (t > top);
o = t % (diff + 1);
return o;
}
/* end source */
I believe this gives a pseudo random in a closed interval with a flat
p.d.f. . Does it look right, besides having some extraneous source? frank
不，它看起来不对 - 在这种情况下，看起来
不是欺骗。除了无用的第三个
参数和愚蠢的无用指针，
函数是错误的。例如，如果要求[42,42]中的值，请考虑会发生什么。
。
-
Eric Sosman
es*****@acm-dot-org.inva lid
No, it does not look right -- and in this case, looks
are not deceiving. Quite aside from the useless third
argument and the silly dabbling with a useless pointer,
the function is wrong. For example, consider what happens
if it is asked for a value in [42,42].
--
Eric Sosman
es*****@acm-dot-org.invalid
Eric Sosman写道：
Eric Sosman wrote:
Frank Silvermann写道：
Frank Silvermann wrote:
[我已经剪掉了95％的代码原来的
>帖子称为外来源。
>这是一个难题。原始海报，知道它是无关的，
>自己没有剪断它。也许下次他会发布一个
>关于三行代码的问题，并将其嵌入
> 尤利西斯的完整文本。
[削减其他95％]
我很难确定相关的内容。虽然我没有使用SWAP宏在原帖上使用SWAP宏，但我很高兴能够在周围处理预期的和非数学的反对意见 br />
被撤销。我相信这会在一个封闭的区间内给出一个伪随机的平坦的p.d.f. 。它看起来是否正确，除了有一些无关的来源？
frank
[I''ve snipped away the 95% of the code that the original
> post called "extraneous source." It''s a puzzlement why
> the original poster, knowing that it was extraneous,
> didn''t snip it himself. Maybe next time he''ll post a
> question about three lines of code, and embed it in the
> complete text of "Ulysses." [snipped away the other 95 percent]
It was hard for me to determine what was relevant. Although I didn''t
use the SWAP macro for on the original post, I was glad to have it
around to deal with the expected and unmathematical objection that they
be reversed. I believe this gives a pseudo random in a closed interval with a flat
p.d.f. . Does it look right, besides having some extraneous source?
frank
不，它看起来不对 - 在这种情况下，看起来不是在欺骗。除了无用的第三个争论和愚蠢的无用指针之外，
功能是错误的。例如，如果在[42,42]中要求输入值，请考虑会发生什么。
No, it does not look right -- and in this case, looks
are not deceiving. Quite aside from the useless third
argument and the silly dabbling with a useless pointer,
the function is wrong. For example, consider what happens
if it is asked for a value in [42,42].
#include< stdio.h>
#include< stdlib.h>
#include< time.h>
#include< math.h>
#define SWAP（m，n）（tmp =（m），（m）=（n），（n）= tmp）
#define ITERATIONS 100
int rand_in_range（int，int）;
int main（无效）
{
int tja，i， j，tmp，m，n;
浮动亩;
/ *所有声明主要需要在这里北方* /
/ *种子计时器* /
srand（时间（NULL））;
/ * call rand * /
j = 0;
m = 40;
n = 3;
if（m> n）SWAP（m，n ）;
printf（范围是[％d，％d] \ n，m，n）;
mu =（浮动）ITERATIONS *（（ float）m +（float）（m - n）/（float）2）;
printf（mu is％f\ n，mu）;
for（i = 0; i< ITERATIONS; i ++）
{
tja = rand_in_range（m，n）;
if（i< 10）printf（return is％d \ n，tja）;
j + = tja;
}
printf（" ; j是％d \ n"，j）;
返回0;
}
int rand_in_range（int m，int n）
{
/ *主要种子srand * /
/ * [m，n]是范围* /
int t，top，diff，o，tmp;
if（m == n）返回m;
if（m > n）SWAP（m，n）;
diff = n - m;
top = RAND_MAX - （RAND_MAX％diff）;
/ * control * /
do t = rand（）; while（t> top）;
o = t％（diff + 1）;
o = o + m;
return o; < br $>
}
/ *结束来源* /
我认为这种情况越来越近了。 mu fiasco显示了为什么它有时很难在动态中删除功能。如果ITERATIONS是10的正值b / b，那么预期的值将是两个输入的小数点移动的平均值。 frank
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#define SWAP(m, n) (tmp = (m), (m) = (n), (n) = tmp)
#define ITERATIONS 100
int rand_in_range(int, int);
int main(void)
{
int tja, i , j, tmp, m, n;
float mu;
/* all declarations in main need to be north of here */
/* seed timer */
srand(time(NULL));
/* call rand */
j = 0;
m = 40;
n = 3;
if (m > n) SWAP(m, n);
printf("range is [%d, %d]\n", m, n);
mu = (float)ITERATIONS *((float)m +(float)(m - n) /(float)2);
printf("mu is %f\n", mu);
for (i=0; i < ITERATIONS; i++)
{
tja = rand_in_range(m, n);
if (i < 10) printf("return is %d\n", tja);
j += tja;
}
printf("j is %d\n", j);
return 0;
}
int rand_in_range(int m, int n)
{
/*seed srand in main */
/* [m, n] is range */
int t, top, diff, o, tmp;
if (m == n) return m;
if (m > n) SWAP(m, n);
diff = n - m;
top = RAND_MAX - (RAND_MAX % diff);
/* control */
do t = rand(); while (t > top);
o = t % (diff + 1) ;
o = o + m;
return o;
}
/* end source */
I think this is getting closer. The mu fiasco shows why it''s sometimes
hard to trim out of a function on the fly. If ITERATIONS is a positive
power of ten, then the expected value is going to be the average of the
two inputs with the decimal point moved. frank
Eric Sosman写道：
Eric Sosman wrote:
Frank Silvermann写道：
Frank Silvermann wrote:
[我已经剪掉了95％原始
[snip]的代码我相信这会在一个封闭的间隔中给出一个伪随机的平面
pdf 。它看起来是否正确，除了有一些无关的来源？
frank
[I''ve snipped away the 95% of the code that the original [snip] I believe this gives a pseudo random in a closed interval with a flat
p.d.f. . Does it look right, besides having some extraneous source?
frank
不，它看起来不对 - 在这种情况下，看起来不是在欺骗。除了无用的第三个争论和愚蠢的无用指针之外，
功能是错误的。例如，如果在[42,42]中要求输入值，请考虑会发生什么。
No, it does not look right -- and in this case, looks
are not deceiving. Quite aside from the useless third
argument and the silly dabbling with a useless pointer,
the function is wrong. For example, consider what happens
if it is asked for a value in [42,42].
/ * rand_in_range.c：贡献者：通常的嫌疑人in clc * /
#include< stdio.h>
#include< stdlib.h>
#include < time.h>
#include< math.h>
#define SWAP（m，n）（tmp =（m），（m）=（ n），（n）= tmp）
#define ITERATIONS 100
int rand_in_range（int，int）;
int main（无效）
{
int tja，i，j，m，n;
/ *主要声明需要在这里的北方* /
/ *种子计时器* /
srand（时间（NULL））;
/ * call rand和计数返回* /
j = 0;
m = 8;
n = 3;
printf（"范围是[％d，％d] \ n"，m，n）;
for（i = 0; i< ITERATIONS; i ++）
{<如果（i <10）printf（return is％d \ n，tja）; br />
j + = tja;
}
printf（" total is％d \ nn，"，j）;
返回0;
}
int rand_in_range（int m，int n）
{
/ * seed srand in main * /
/ * [m，n]是范围* /
int t，top，diff，p，tmp;
如果（m == n）返回m;
if（m> n）SWAP（m，n）;
diff = n - m;
top = RAND_MAX - （RAND_MAX％（diff + 1）+ 1）;
/ * control * /
do t = rand（）; while（t> top）;
p = t％（diff + 1）;
p = p + m;
返回p; < br $>
}
/ *结束来源* /
我认为这是对的。 frank
/* rand_in_range.c : contributors: usual suspects in clc */
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <math.h>
#define SWAP(m, n) (tmp = (m), (m) = (n), (n) = tmp)
#define ITERATIONS 100
int rand_in_range(int, int);
int main(void)
{
int tja, i , j, m, n;
/* all declarations in main need to be north of here */
/* seed timer */
srand(time(NULL));
/* call rand and count returns */
j = 0;
m = 8;
n = 3;
printf("range is [%d, %d]\n", m, n);
for (i=0; i < ITERATIONS; i++)
{
tja = rand_in_range(m, n);
if (i < 10) printf("return is %d\n", tja);
j += tja;
}
printf("total is %d\n", j);
return 0;
}
int rand_in_range(int m, int n)
{
/*seed srand in main */
/* [m, n] is range */
int t, top, diff, p, tmp;
if (m == n) return m;
if (m > n) SWAP(m, n);
diff = n - m;
top = RAND_MAX - (RAND_MAX % (diff + 1) + 1);
/* control */
do t = rand(); while (t > top);
p = t % (diff + 1) ;
p = p + m;
return p;
}
/* end source */
I think this is right. frank
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