进入a b 多少努力p, q 使p*q == a && p < q && p >= b
直接大整数分解 然后dfs所有可能的解决方案劫持
#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;
}
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