任何人对此都有任何想法..... [英] Anyone have any ideas on this one.....

问题描述

我正在尝试编写一个程序，它将在1到50之间生成100个随机数

。使用这些数字，我想生成一个列表，这将是

告诉随机数在1-5,6-10,11-15之间，

16-20，......和46-50。最后，我想打印出结果作为

直方图。我把它写出来并认为它会是这样的。


1-5（11）***********

6-10（8）********

11-15（12）************

16 -20（9）*********

21-25（10）**********

26-30（ 11）***********

31-35（7）*******

36-40（8）** ******

41-45（13）*************

46-50（11）*** ********


我正在努力学习如何做到这一点而且我被卡住了。任何建议都会有帮助。
帮助。甚至一些可能的教程也会有所帮助。

-

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I am trying to write a program that will generate 100 random numbers
between 1 and 50. Using these numbers, I want to generate a list that will
tell the number of random numbers that fell between 1-5, 6-10, 11-15,
16-20, ... , and 46-50. Finally,I would like to print out the results as a
histogram. I have written it out and would think it would look like this.

1-5 (11) ***********
6-10 (8) ********
11-15 (12) ************
16-20 (9) *********
21-25 (10) **********
26-30 (11) ***********
31-35 (7) *******
36-40 (8) ********
41-45 (13) *************
46-50 (11) ***********

I am trying to learn how to do this and I am stuck. Any suggestions would
help. Even some possible tutorials to chek out would help.
--
Message posted using http://www.talkaboutprogramming.com/group/comp.lang.c/
More information at http://www.talkaboutprogramming.com/faq.html

推荐答案

" joebenjamin" < gu ********* @ hotmail.comwrites：

"joebenjamin" <gu*********@hotmail.comwrites:

我正在尝试编写一个可生成100个随机数的程序

在1到50之间。使用这些数字，我想生成一个列表，它将

告知1-5,6-10,11-15之间的随机数，

16-20，......和46-50。最后，我想打印出结果作为

直方图。我把它写出来并认为它会是这样的。


1-5（11）***********

6-10（8）********

11-15（12）************

16 -20（9）*********

21-25（10）**********

26-30（ 11）***********

31-35（7）*******

36-40（8）** ******

41-45（13）*************

46-50（11）*** ********


我正在努力学习如何做到这一点而且我被卡住了。任何建议都会有帮助。
帮助。甚至一些可能的教程也会有所帮助。

I am trying to write a program that will generate 100 random numbers
between 1 and 50. Using these numbers, I want to generate a list that will
tell the number of random numbers that fell between 1-5, 6-10, 11-15,
16-20, ... , and 46-50. Finally,I would like to print out the results as a
histogram. I have written it out and would think it would look like this.

1-5 (11) ***********
6-10 (8) ********
11-15 (12) ************
16-20 (9) *********
21-25 (10) **********
26-30 (11) ***********
31-35 (7) *******
36-40 (8) ********
41-45 (13) *************
46-50 (11) ***********

I am trying to learn how to do this and I am stuck. Any suggestions would
help. Even some possible tutorials to chek out would help.

要从set [1,50]生成随机数整数，您可以使用：" number

= rand（）％50 + 1;" （记得初始化伪随机数

生成器，例如srand（time（0）））。


要获得桶号，你可以使用 bucket =（number - 1）/ 5;" （所以在

中，最好使用

" number = rand（）％50;"从set [0,49]生成随机整数然后计算桶号使用bucket =

number / 5;。


然后你增加给定的桶，即：桶[桶] ++" ;.当然，铲斗首先要归零。当结果准备就绪时，你使用两个嵌套的for循环打印

结果。


-

祝你好运，_ _

.o。 | Serenly Enlightened Majesty of o''，=。/`o

..o |计算机科学，Michalmina86 Nazarewicz（oo）

ooo + - < mina86 * tlen.pl> ---< jid：mina86 * chrome.pl> - ooO - （_） - Ooo--

To generate random number integer from set [1, 50] you can use: "number
= rand() % 50 + 1;" (remember to initialise pseudo-random number
generator with for instance srand(time(0))).

To get bucket number you can use "bucket = (number - 1) / 5;" (so in
fact it is better to generate random integer from set [0, 49] using
"number = rand() % 50;" and then calculate bucket number using "bucket =
number / 5;").

Then you increment given bucket, ie.: "buckets[bucket]++". Buckets have
to be zeroed first of course. And when results are ready you print the
results using two nested for loops.

--
Best regards, _ _
.o. | Liege of Serenly Enlightened Majesty of o'' \,=./ `o
..o | Computer Science, Michal "mina86" Nazarewicz (o o)
ooo +--<mina86*tlen.pl>---<jid:mina86*chrome.pl>--ooO--(_)--Ooo--

Michal Nazarewicz说：

Michal Nazarewicz said:

" joebenjamin" < gu ********* @ hotmail.comwrites：

"joebenjamin" <gu*********@hotmail.comwrites:

>我正在尝试编写一个可生成100个随机数的程序将告诉1-5,16-10之间的随机数，11-15,16- 20，...，和46-50。最后，我想打印出
结果作为直方图。我已经把它写出来并认为它会是这样的。

1-5（11）***********
6-10 （8）********
11-15（12）************
16-20（9）******* **
21-25（10）**********
26-30（11）***********
31-35 （7）*******
36-40（8）********
41-45（13）************ *
46-50（11）***********

我正在努力学习如何做到这一点而且我被困住了。任何建议都会有所帮助。甚至一些可能的教程也会有所帮助。

>I am trying to write a program that will generate 100 random numbers
between 1 and 50. Using these numbers, I want to generate a list that
will tell the number of random numbers that fell between 1-5, 6-10,
11-15, 16-20, ... , and 46-50. Finally,I would like to print out the
results as a histogram. I have written it out and would think it
would look like this.

1-5 (11) ***********
6-10 (8) ********
11-15 (12) ************
16-20 (9) *********
21-25 (10) **********
26-30 (11) ***********
31-35 (7) *******
36-40 (8) ********
41-45 (13) *************
46-50 (11) ***********

I am trying to learn how to do this and I am stuck. Any suggestions
would help. Even some possible tutorials to chek out would help.

要从set [1,50]生成随机数整数，您可以使用：

" number = rand（）％50 + 1;"

To generate random number integer from set [1, 50] you can use:
"number = rand() % 50 + 1;"

更好：int number = 50 *（rand（）/（RAND_MAX + 1.0））;

Better: int number = 50 * (rand() / (RAND_MAX + 1.0));

（记得使用例如srand（time（0））初始化伪随机

数字生成器）。

(remember to initialise pseudo-random
number generator with for instance srand(time(0))).

如果他想要重复性，他最好使用常数。如果他b $ b想要不可预测性（尽可能实用），以下函数

（基于Lawrence Kirby编写的代码）可以更好地完成

利用所有可用的熵。


#include< stddef.h>

#include< time.h>

#include< limits.h>


/ *用法：srand（time_seed（））; * /


/ *根据当前时间选择并返回初始随机种子。

基于Lawrence Kirby的代码< fr ** @ genesis .demon.co.uk取代。 * /

未签名

time_seed（无效）

{

time_t timeval;

unsigned char * ptr;

未签名的种子;

size_t i;


timeval = time（NULL）;

ptr =（unsigned char *）& timeval;


seed = 0;

for（i = 0; i< sizeof timeval; i ++）

seed = seed *（UCHAR_MAX + 2U）+ ptr [i];


返回种子;

}

If he wants repeatability, he''d be better off using a constant. If he
wants unpredictability (as far as is practical), the following function
(based on code written by Lawrence Kirby) does a better job of
utilising all the available entropy.

#include <stddef.h>
#include <time.h>
#include <limits.h>

/* Usage: srand (time_seed ()); */

/* Choose and return an initial random seed based on the current time.
Based on code by Lawrence Kirby <fr**@genesis.demon.co.uk>. */
unsigned
time_seed (void)
{
time_t timeval;
unsigned char *ptr;
unsigned seed;
size_t i;

timeval = time (NULL);
ptr = (unsigned char *) &timeval;

seed = 0;
for (i = 0; i < sizeof timeval; i++)
seed = seed * (UCHAR_MAX + 2U) + ptr[i];

return seed;
}

>

要获得桶号，您可以使用bucket =（number - 1）/ 5;; （所以在

中，最好使用

" number = rand（）％50;"从set [0,49]生成随机整数然后计算桶号使用桶

=数字/ 5;。


然后你增加给定桶，即：桶[桶] ++" ;.铲斗

首先将
归零。当结果准备就绪时，使用两个嵌套的for循环打印结果

。

>
To get bucket number you can use "bucket = (number - 1) / 5;" (so in
fact it is better to generate random integer from set [0, 49] using
"number = rand() % 50;" and then calculate bucket number using "bucket
= number / 5;").

Then you increment given bucket, ie.: "buckets[bucket]++". Buckets
have
to be zeroed first of course. And when results are ready you print
the results using two nested for loops.

-

Richard Heathfield< http://www.cpax.org.uk>

电子邮件：-www。 + rjh @

谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>

Usenet是一个奇怪的放置" - dmr 1999年7月29日

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

试试这个：


#include< stdio.h>

#include< stdlib.h>

#include< time.h>


int main（无效）

{

int matrix [10];

int i;


int x;


srand（time（NULL））;

for（i = 0; i< 10; i ++）

matrix [i] = 0;


for（i = 0; i< 100; i ++）{

x = rand（）％50 + 1;


if（x< ; 6）

matrix [0] ++;

else if（x< 11）

matrix [1] ++;

else if（x< 16）

matrix [2] ++;

else if（x< 21）

matrix [3] ++;

else if（x< 26）

matrix [4] ++;

else if（x< ; 31）

matrix [5] ++;

else if（x <36）

matrix [6] ++;

else if（x< 41）

matrix [7] ++;

else if（x< 46）

matrix [8] ++;

else

matrix [9] ++;

}

printf （QUOT; 1-5 \t\t（％3d）"，matrix [0]）;

for（i = 0; i< matrix [0]; i ++）

printf（" *"）;

printf（" \ n"）;


printf（" 6-10 \\ \\ t \\ t（％3d）"，matrix [1]）;

for（i = 0; i< matrix [1]; i ++）

printf （" *"）;

printf（" \ n"）;


返回0;

}

Try this:

#include<stdio.h>
#include<stdlib.h>
#include<time.h>

int main(void)
{
int matrix[10];
int i;

int x;

srand(time(NULL));
for(i=0;i<10;i++)
matrix[i]=0;

for(i=0;i<100;i++){
x=rand()%50 + 1;

if(x<6)
matrix[0]++;
else if(x<11)
matrix[1]++;
else if(x<16)
matrix[2]++;
else if(x<21)
matrix[3]++;
else if(x<26)
matrix[4]++;
else if(x<31)
matrix[5]++;
else if(x<36)
matrix[6]++;
else if(x<41)
matrix[7]++;
else if(x<46)
matrix[8]++;
else
matrix[9]++;
}
printf("1-5 \t\t ( %3d ) ",matrix[0]);
for(i=0;i<matrix[0];i++)
printf("*");
printf("\n");

printf("6-10 \t\t ( %3d ) ",matrix[1]);
for(i=0;i<matrix[1];i++)
printf("*");
printf("\n");

return 0;
}

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