使用循环数组生成唯一的随机编号 [英] Generating Unique Random Numbers in an Array using Loop
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问题描述
所以现在的问题是建立一个[5] [5]表中,每个包含独特的人数从1-100(的没有重复的)
So the question is to develop a [5][5] table, each containing unique numbers from 1-100 (no duplicates)
所以这里就是我想出了:
so here's what I came up with:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main()
{
int outerLoop;
int innerLoop;
int board[5][5]; /* Array Name And Size*/
/* seeds the random number generator*/
srand(time(NULL));
int number;
number = rand() % 101;
/* Start a loop to generate a random integer between 1 and 100 and
assign it into the array. The loop runs 25 times*/
for ( outerLoop = 0 ; outerLoop <= 25 ; outerLoop++ ) /* loop 25 times*/
{
for ( innerLoop = 0 ; innerLoop <= 4 ; innerLoop++ ) /* <=4 due to 5
columns*/
{
board[outerLoop][innerLoop] = rand() % 100 + 1;
}
printf( "%d \n", board[outerLoop][innerLoop] );
}
所以我pretty多被困here.I'm不能肯定这一点:
So I pretty much got stuck here.I'm not really sure about this:
board[outerLoop][innerLoop] = rand() % 100 + 1;
我只是做它:/任何想法的家伙
I simply made it up :/ Any idea guys?
推荐答案
你想要的是一个洗牌的算法
What you want is a shuffle algorithm
要获得独一无二的#秒25的元素的数组从1到100;只创建与数字1..100 100元件阵列,从100池中洗牌一号25,并用第1 25
To get your 25 element array of unique #s from 1 to 100; just create a 100 element array with the numbers 1..100, shuffle the 1st 25 from the pool of 100, and use the 1st 25.
$ cat test.c
#include <stdio.h>
#include <stdlib.h>
void shuffle(int *array, size_t array_size, size_t shuff_size)
{
if (array_size > 1)
{
size_t i;
for (i = 0; i < shuff_size - 1; i++)
{
size_t j = i + rand() / (RAND_MAX / (array_size - i) + 1);
int t = array[j];
array[j] = array[i];
array[i] = t;
}
}
}
int main(int argc, char * argv[]) {
int a[100];
int b[5][5];
int i,j,k=0;
for(i=0; i<100;++i)
a[i]=i;
shuffle(a,100,25);
for(i=0;i<5;++i)
for(j=0;j<5;++j) {
b[i][j] = a[k++];
printf("%d ",b[i][j]);
}
printf("\n");
}
$ gcc -o test test.c
$ ./test
0 14 76 47 55 25 10 70 7 94 44 57 85 16 18 60 72 17 49 24 53 75 67 9 19
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