The most divisor problem, find the number of divisors of a number

A mathematical analysis

 two-correlation principle

https://m.baike.so.com/doc/3567228-3751425.html

Three practical exercises

most divisor problem

Problem description: The divisor of a positive integer x is a positive integer that can divide x evenly. The number of divisors of a positive integer x is recorded as div(x). For example, 1, 2, 3, and 10 are all divisors of the positive integer 10, and div(10)=4. Suppose a and b are two positive integers, a<=b, find the number x between a and b with the largest number of divisors.

The following demonstrates the implementation in C language

1. A function that generates prime numbers

void primes(){
	bool get [MAXP+1];
	for(int i=2;i<=MAXP;i++)
		get[i]=ture;
	for(i=2;i<=MAXP;i++)
		if(get[i]){
			int j=i+1;
			while(j<=MAXP){
				get[j]=false;
				j+=i;
			}
		}
	for(int ii=2,j=0;ii<=MAXP;ii++)
		if(get[ii]){
			prim[++j]=ii;
		}

2. Search for the function with the most divisors

void search(int from,int tot,int num,int low,int up){
	if(num>=1)
		if((tot>max)||((tot==max)&&(num<numb))){
			max=tot;
			numb=num;
		}
	if((low==up)&&(low>num))
		search(from,tot*2,num*low,1,1);
	for(int i=from;i<=PCOUNT;i++){
		if(prim[i]>up)
			return;
		else{
			int j=prim[i],x=low1,y=up,n=num,t=tot,m=1;
			while(ture){
				m++;
				t+=tot;
				x/=j;
				y/=j;
				if(x==y)
					break;
				n*=j;
				search(i+1,t,n,x+1,y);}
		}
		m=1<<m;
		if(tot<max/m)
			return;
	}
}

3. Main function

int main(){
    primes();
    cin>>l>>u;
    if((l==1)&&(u==1)){
        max=1;
        numb=1;
    }
    else{
        max=2;
        numb=1;
        search(1,1,1,l,u);
		}
	cout<<max<<end1;
		return 0;
	}