SPOJ 34096 Counting Divisors (general)

考虑函数 $f(x)={\sigma }_0(x^k)$ 在质数和质数次幂处的值。

考虑 $f(p)$，容易发现 $f(p)={\sigma }_0(p^k)=k+1$。
考虑 $f(p^c)$，容易发现 $f(p^c)={\sigma }_0((p^c{)}^k)={\sigma }_0(p^{ck})=ck+1$。
其中 $p$ 为任意质数。

于是直接套用模板即可。

代码：

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const long long N = 1e10;
const int MX = 1e5;
int T;
unsigned long long n,K;
int lim;
int vis[MX + 5],cnt,prime[MX + 5];
int tot,le[MX + 5],ge[MX + 5];
inline int &id(long long x)
{
    return x <= lim ? le[x] : ge[n / x];
}
unsigned long long lis[2 * MX + 5];
unsigned long long G[2 * MX + 5],Fprime[2 * MX + 5];
unsigned long long F(int k,unsigned long long n)
{
    if(n < (unsigned long long)prime[k] || n <= 1)
        return 0;
    unsigned long long ret = Fprime[id(n)] - (k - 1) * (K + 1);
    for(register int i = k;i <= cnt && (unsigned long long)prime[i] * prime[i] <= n;++i)
    {
        unsigned long long pw = prime[i],pw2 = (unsigned long long)prime[i] * prime[i];
        for(register int c = 1;pw2 <= n;++c,pw = pw2,pw2 *= prime[i])
            ret += (c * K + 1) * F(i + 1,n / pw) + ((c + 1) * K + 1);
    }
    return ret;
}
int main()
{
    for(register int i = 2;i <= MX;++i)
    {
        if(!vis[i])
            prime[++cnt] = i;
        for(register int j = 1;j <= cnt && i * prime[j] <= MX;++j)
        {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j]))
                break;
        }
    }
    scanf("%d",&T);
    for(;T;--T)
    {
        scanf("%llu%llu",&n,&K),lim = sqrt(n),tot = 0;
        for(unsigned long long l = 1,r;l <= n;l = r + 1)
        {
            r = n / (n / l);
            lis[id(n / l) = ++tot] = n / l;
            G[tot] = n / l - 1;
        }
        for(register int k = 1;k <= cnt;++k)
        {
            int p = prime[k];
            unsigned long long s = (unsigned long long)p * p;
            for(register int i = 1;lis[i] >= s && i <= tot;++i)
                G[i] -= G[id(lis[i] / p)] - (k - 1);
        }
        for(register int i = 1;i <= tot;++i)
            Fprime[i] = (K + 1) * G[i];
        printf("%llu\n",F(1,n) + 1);
    }
}