LuoGu: SP19975 APS2 - Amazing Prime Sequence (hard)
SPOJ: DIVFACT4 - Divisors of factorial (extreme)
对于 $i \le \sqrt n$ 的质数，统计 $[i + 1, n]$ 有多少个数最小质因数为 $i$ 。这可以在 Min_25 第一步过程中计算。对于更大的质数，贡献只有自己，考虑计算所有质数的和，这也可以用 Min_25 第一步计算。
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 = 3e6 + 10;
bool vis[N];
LL n, w[N];
int B, tot;
ULL f[N], g[N], ps[N], ans;
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;
}
}
for (int i = 1; i <= cnt; ++i)
ps[i] = ps[i - 1] + p[i];
}
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;
g[i] = (w[i] & 1 ? (w[i] + 1) / 2 * w[i] : w[i] / 2 * (w[i] + 1)) - 1;
}
for (int i = 1; i <= cnt && p[i] <= B; ++i)
for (int j = 1; j <= tot && w[j] >= (LL)p[i] * p[i]; ++j)
{
int t = id(w[j] / p[i]);
if (j == 1) ans += (ULL)p[i] * (f[t] - (i - 1));
f[j] -= f[t] - (i - 1);
g[j] -= (ULL)p[i] * (g[t] - ps[i - 1]);
}
}
int main ()
{
init();
int T; read(T);
while (T--)
{
read(n); B = sqrt(n);
tot = 0; ans = 0;
for (LL l = 1, r; l <= n; l = r + 1)
{
r = n / (n / l);
w[id(n / l) = ++tot] = n / l;
}
solve();
write(ans + g[1]), puts("");
}
return 0;
}
|