wish help you to fix your issue I want to generate 7 random numbers from 0 to 39 and store them in one dimensional array. I must make sure each number is different (can't have two 7s for example). , OK, here's my contribution: code :
#include <iostream>
#include <algorithm>
#include <random>
#include <vector>
int main()
{
std::vector <int> v (40);
std::generate (v.begin (), v.end (), [n = 0] () mutable { return n++; });
std::random_device rd;
std::mt19937 g (rd ());
std::shuffle (v.begin (), v.end (), g);
for (int i = 0; i < 7; ++i)
std::cout << v [i] << '\n';
}
5
39
10
17
36
11
31
std::iota (v.begin (), v.end (), 0);
Share :

Selecting four random numbers from nine without repeats
By : Kamal Bhattacharjee
Date : March 29 2020, 07:55 AM
wish helps you I think you need to either rename your check boxes or use square brackets: code :
[1].CheckState = 1 'for example.
If l(ln) = 1 & n(lm) = 1 Then 1.CheckState = 1
If l(ln) = 1 And n(lm) = 1 Then [1].CheckState = 1
[1].CheckState = 1 And (l(ln) = 1) And (n(lm) = 1)

Generate random numbers without repeats
By : user3190681
Date : March 29 2020, 07:55 AM
hop of those help? Say you want to generate 1,000 unique random numbers and present them to some code one at a time. When you exhaust those numbers, you want to present the same numbers again, but in a different sequence. To generate the numbers, use a hash table. In C#, it would look like this: code :
const int MaxNumbers = 1000;
HashSet<int> uniqueNumbers = new HashSet<int>();
Random rnd = new Random();
// generate random numbers until there are 1000 in the hash table
while (uniqueNumbers.Count < MaxNumbers)
{
uniqueNumbers.Add(rnd.Next());
}
// At this point you have 1,000 unique random numbers
// Put them in an array
int[] myNumbers = uniqueNumbers.ToArray();
int ixCurrent = 0;
int GetNextNumber()
{
if (ixCurrent < MaxNumbers)
{
++ixCurrent;
return myNumbers[ixCurrent1];
}
// out of numbers. Shuffle the array and return the first one.
for (int i = MaxNumbers1; i > 0; i)
{
int j = rnd.Next(i+1);
int temp = myNumbers[i];
myNumbers[i] = myNumbers[j];
myNumbers[j] = temp;
}
ixCurrent = 1;
return myNumbers[0];
}

Is there a good way to generate random data in Lua with no repeats?
By : Sandeep Tete
Date : March 29 2020, 07:55 AM
This might help you Many people often face this same problem, here I provide the solution that helps probably about 99% of them: code :
math.randomseed(os.time())
local left = {}
for i = 1,10 do
left[i] = i
end
local function Random()
return table.remove(left, math.random(#left))
end
for i = 1,10 do
print(Random())
end

C  Generate random sequence with no repeats without shuffling
By : Fördős Gyula Gábor
Date : March 29 2020, 07:55 AM
Hope this helps This can be done in O(n) time with two lists, one with the number (initialy) in order, and one in the resulting order. You start with n elements in order in your source list. Then you select a random number mod n. That gives you the next element, which you place in the destination list. code :
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define LEN 10
int main()
{
int a[LEN], b[LEN];
int i, val;
int count = LEN;
srand(time(NULL));
for (i=0;i<LEN;i++) {
a[i]=i+1;
}
for (i=0;i<LEN;i++) {
val = rand() % count;
b[i] = a[val];
a[val] = a[count1];
count;
}
for (i=0;i<LEN;i++) {
printf("%d ", b[i]);
}
printf("\n");
return 0;
}
int a[LEN];
int i, val, tmp;
srand(time(NULL));
for (i=0;i<LEN;i++) {
a[i]=i+1;
}
for (i=0;i<LEN1;i++) {
val = (rand() % (LEN  1  i)) + i + 1;
tmp = a[i];
a[i] = a[val];
a[val] = tmp;
}
for (i=0;i<LEN;i++) {
printf("%d ", a[i]);
}
printf("\n");

How to generate a bunch of lists of random numbers which have no repeats but in a certain range with Python?
By : user3624565
Date : March 29 2020, 07:55 AM
will be helpful for those in need I actually need to generate a bunch of lists of random numbers in the range of 1 to 5. I know how to generate a single list of random numbers in the range of 1 to 5 with shuffle module, but what if I want a bunch of such stuffs? I have no idea to use loop ,is there anyone can help? Many appreciates~ code :
import random
f = []
for i in range(5):
f.append(random.sample(range(5), 5))
print(f)
x = [random.sample(range(5), 5) for i in range(5)]
print(x)
[[3, 0, 1, 4, 2], [4, 2, 0, 3, 1], [4, 0, 1, 3, 2], [3, 0, 1, 2, 4], [4, 3, 2, 0, 1]]

