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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

SP34096 DIVCNTK - Counting Divisors (general)

LuoGu: SP34096 DIVCNTK - Counting Divisors (general)

SPOJ: (DIVCNTK - Counting Divisors (general))

记 $F(x) = \sigma(x ^ k)$ ,可以发现这是一个积性函数。在质数上的值有 $F(p) = k + 1$ ,提取 $k + 1$ ,计算 $F(p) = 1$ ,是一个完全积性函数。那么可以直接用 Min_25 筛完成。

查看代码
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#pragma GCC optimize ("O2")
#include <cstdio>
#include <cmath>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 1) + (x << 3) + c - '0';
flag && (x = ~x + 1);
}
template <class Type, class ...rest>
void read (Type &x, rest &...y) { read(x), read(y...); }
template <class Type>
void write (Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar('0' + x % 10);
}
typedef long long LL;
typedef unsigned long long ULL;
const int N = 2e5 + 10;
LL n, m, w[N];
bool vis[N];
int B, tot;
ULL f[N];
int cnt, p[N], id1[N], id2[N];
void init ()
{
vis[1] = true;
for (int i = 2; i < N; ++i)
{
if (!vis[i]) p[++cnt] = i;
for (int j = 1; j <= cnt && i * p[j] < N; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0) break;
}
}
}
int &id (LL x) { return x <= B ? id1[x] : id2[n / x]; }
void solve ()
{
for (int i = 1; i <= tot; ++i)
f[i] = (w[i] - 1);
for (int i = 1; i <= cnt; ++i)
for (int j = 1; j <= tot && w[j] >= (LL)p[i] * p[i]; ++j)
f[j] -= f[id(w[j] / p[i])] - (i - 1);
}
ULL S (LL a, int b)
{
if (p[b] >= a) return 0;
ULL res = (f[id(a)] - b * (m + 1));
for (int i = b + 1; i <= cnt && (LL)p[i] * p[i] <= a; ++i)
{
LL t = p[i];
for (int k = 1; t <= a; ++k, t *= p[i])
res += (m * k + 1) * (S(a / t, i) + (k > 1));
}
return res;
}
int main ()
{
init();
int T; read(T);
while (T--)
{
tot = 0;
read(n, m); B = sqrt(n);
for (LL l = 1, r; l <= n; l = r + 1)
{
r = n / (n / l);
w[id(n / l) = ++tot] = n / l;
}
solve();
for (int i = 1; i <= tot; ++i)
f[i] *= m + 1;
write((S(n, 0) + 1)), puts("");
}
return 0;
}