P6055 [RC-02] GCD
易得答案为
$$
\sum _ {i = 1} ^ n \mu (i) \lfloor \frac n i \rfloor ^ 3
$$
用杜教筛筛 $\mu$ 。
查看代码
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| #include <cstdio>
#include <map>
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;
const int N = 2e6 + 10, mod = 998244353;
bool vis[N];
int n, cnt, p[N], f[N];
map <int, int> F;
void init ()
{
f[1] = 1;
vis[1] = true;
for (int i = 2; i < N; ++i)
{
if (!vis[i]) f[p[++cnt] = i] = -1;
for (int j = 1; j <= cnt && i * p[j] < N; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0) break;
f[i * p[j]] = -f[i];
}
}
for (int i = 2; i < N; ++i) f[i] += f[i - 1];
}
int calc (int x)
{
if (x < N) return f[x];
if (F.count(x)) return F[x];
int res = 1;
for (int l = 2, r; l <= x; l = r + 1)
{
r = x / (x / l);
res -= (r - l + 1) * calc(x / l);
}
return F[x] = res;
}
int a3 (int x) { return (LL)x * x % mod * x % mod; }
int main ()
{
init();
read(n);
int res = 0;
for (int l = 1, r; l <= n; l = r + 1)
{
r = n / (n / l);
res = ((LL)a3(n / l) * (calc(r) - calc(l - 1)) + res) % mod;
}
write((res + mod) % mod);
return 0;
}
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