输入a b 求有多少对p, q 使得p*q == a && p < q && p >= b

直接大整数分解 然后dfs搜出所有可能的解

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long LL;
const int Times = 25;
LL factor[100], f[100];
int l, ll, ans, num[100];		
LL a, b;

LL gcd(LL a, LL b)
{
	return b ? gcd(b, a%b):a;
}
LL add_mod(LL a, LL b, LL n)
{
    LL ans = 0;
    while(b)
    {
        if(b&1)
            ans = (ans + a)%n;
        b >>= 1;
        a = (a<<1)%n;
    }
    return ans;
}
LL pow_mod(LL a, LL m, LL n)
{
    LL ans = 1;
    while(m)
    {
        if(m&1)
            ans = add_mod(ans, a, n);
        m >>= 1;
        a = add_mod(a, a, n);
    }
    return ans;
}
bool Witness(LL a, LL n)
{
    int j = 0;
    LL m = n-1;
    while(!(m&1))
    {
        j++;
        m >>= 1;
    }
    LL x = pow_mod(a, m, n);
    if(x == 1 || x == n-1)
        return true;
    while(j--)
    {
        x = add_mod(x, x, n);
        if(x == n-1)
            return true;
    }
    return false;
}
bool Miller_Rabin(LL n)
{
    if(n < 2)
        return false;
    if(n == 2)
        return true;
    if(!(n&1))
        return false;
    for(int i = 0; i < Times; i++)
    {
        LL a = rand()%(n-1)+1;
        if(!Witness(a, n))
            return false;
    }
    return true;
}
LL Pollard_rho(LL n, LL c)
{
	LL i = 1, x = rand()%(n-1)+1, y = x, k = 2, d;
	//srand(time(NULL));
	while(true)
	{
		i++;
		x = (add_mod(x,x,n)+c)%n;
		d = gcd(y-x,n);
		if(d > 1 && d < n)
			return d;
		if(y == x)
			return n;
		if(i == k)
		{
			y = x;
			k <<= 1;
		}
	}
}
void get_fact(LL n, LL k)
{
	if(n == 1)
		return;
	if(Miller_Rabin(n))
	{
		factor[l++] = n;
		return;
	}
	LL p = n;
	while(p >= n)
	{
		p = Pollard_rho(p, k--);
	}
	get_fact(p, k);
	get_fact(n/p, k);
}

void dfs(LL x, int p, LL m)
{
	if(x > m)
		return;
	if(p == ll)
	{
		if(x >= b && a/x > x)
			ans++;
		//printf("%lld\n", x);
		return;
	}
	LL y = 1;
	for(int i = 0; i <= num[p]; i++)
	{
		dfs(x*y, p+1, m);
		y *= f[p];
	}
}
int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{

		scanf("%lld %lld", &a, &b);
		LL m = sqrt(a+0.5);
		ans = 0;
		l = 0;
		get_fact(a, 120);
		sort(factor, factor+l);
		f[0] = factor[0];
		num[0] = 1;
		ll = 1;
		for(int i = 1; i < l; i++)
		{
			if(factor[i] != factor[i-1])
			{
				ll++;
				f[ll-1] = factor[i];
				num[ll-1] = 0;
			}
			num[ll-1]++;
		}
		//for(int i = 0; i < ll; i++)
		//	printf("%lld %d\n", f[i], num[i]);
		dfs(1, 0, m);
		printf("Case %d: %d\n", cas++, ans);
	}
	return 0;
}

Light OJ 1341 Aladdin and the Flying Carpet Pollard_rho整数分解+DFS