Randomly generated m numbers between 0 and N

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Randomly generated m numbers between 0 and N

How to generate a non-repeat number of m between 0 and N with random numbers

1, the most direct way is to randomly generate a number between 0 to N, to determine whether this number has been selected, if not previously selected, then select, if previously selected, discard

void Common (int n,int m)
{
	int * randnum= (int *) malloc (n*sizeof (int));
	memset (randnum,0,n*sizeof (int));   Place all n positions at 0
	Srand (Time (NULL));
	while (m)
	{
		int cur=rand ()%n;
		if (randnum[cur]==0)   //Make a judgment, select and output
		{
			cout<<cur<<endl;
			if the current number is not selected Randnum[cur]=1;
			m--
		}
	}
	Free (randnum);

}

This approach is easy to understand, but requires extra space to ensure that the number of removed is not repeated, so we have a simpler way, the answer is yes

2, first on the code, after the explanation

void Mrand (int n, int m)
{

	Srand (Time (NULL));
	for (int i=0;i<n;i++)
	{	
		if (rand ()% (n-i) <m)
		{
			cout<<i<<endl;
			m--
		}}
		
	}

The above code is concise, but not easy to understand, we next explain

The first is a loop, which ensures that the number of outputs is not duplicated, because every time I am not the same

The second number is M, in which the rand ()% (n-i) <m is used to determine if the number is less than M, and M minus 1 if the condition is met until 0, which means the number of M has been taken.

Again is how to ensure that the number of M is equal to the probability of

In the first cycle of i=0, n-i=n, then the random number is generated by the 0-n-1 between the random number, then at the moment 0 is taken to the probability of m/n-1
I=1,n-i=n-1 in the second cycle, the random number generates a random number between 0-n-2, at which point the probability of 1 being taken is related to the 0 of the previous cycle. Let's say that in the last loop, not taken, then the probability of the 1 is m/n-2; assuming that the last cycle has been taken, then the probability of taking 1 is m-1/n-2, so the probability of this being taken in general is (1-m/n-1) * (m/n-2) + (m/n-1) * (m-1/ N-2), the result of the final Tong merging is the same as the probability of m/n-1 and the first time
Similarly, the probability of I being taken is also m/n-1 in the first cycle.

So the m number is equal to the probability.