P1447 [NOI2010] 能量采集
求
$$
\begin {aligned}
& \sum _ {i = 1} ^ n \sum _ {j = 1} ^ m 2(i, j) - 1 \\
= & 2\sum _ {d = 1} ^ n \varphi (d) \lfloor \frac n d \rfloor \lfloor \frac m d \rfloor - nm
\end {aligned}
$$
还是数论分块。
查看代码
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| #include <cstdio>
#include <algorithm>
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 << 3) + (x << 1) + c - '0';
if (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)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
typedef long long LL;
const int N = 1e5 + 10;
int n, m;
LL f[N];
int cnt, p[N];
bool vis[N];
void init()
{
f[1] = 1;
vis[1] = true;
for (int i = 2; i < N; ++i)
{
if (!vis[i]) p[++cnt] = i, f[i] = i - 1;
for (int j = 1; i * p[j] < N; ++j)
{
vis[i * p[j]] = true;
if (i % p[j] == 0)
{ f[i * p[j]] = f[i] * p[j]; break; }
f[i * p[j]] = f[i] * f[p[j]];
}
}
for (int i = 2; i < N; ++i) f[i] += f[i - 1];
}
int main ()
{
init();
read(n, m);
if (n > m) swap(n, m);
LL res = 0;
for (int l = 1, r; l <= n; l = r + 1)
{
r = min(n / (n / l), m / (m / l));
res += (LL)(f[r] - f[l - 1]) * (n / l) * (m / l);
}
res = res * 2 - (LL)n * m;
write(res);
return 0;
}
|