LuoGu: SP21615 NAJPWG - Playing with GCD
SPOJ: NAJPWG - Playing with GCD
莫反当然可以搞，但是加上数论分块后 $O(\sqrt n)$ 的单次询问的复杂度比较卡。
$$
\begin {array}{c}
& \displaystyle \sum _ {i = 1} ^ n \sum _ {j = i} ^ n [(i, j) > 1] \\
\Longrightarrow & \displaystyle \sum _ {i = 1} ^ n \sum _ {j = 1} ^ {i - 1} [(i, j) > 1] \\
\end {array}
$$
也就是交换了 $i$ $j$ 的枚举顺序。
注意到
$$
\displaystyle \sum _ {i = 1} ^ {j - 1} [(i, j) = 1] = \varphi (j)
$$
故答案为
$$
\sum _ {i = 1} ^ n i - \varphi (i)
$$
预处理后，$O(1)$ 后回答。
查看代码
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| #include <cstdio>
using namespace std;
const int N = 1e5 + 10;
bool vis[N];
int cnt, primes[N], low[N], phi[N], s[N];
void init()
{
phi[1] = 1;
low[1] = 1;
for (int i = 2; i < N; i++)
{
if (!vis[i])
{
primes[++cnt] = i;
phi[i] = i - 1;
low[i] = i;
}
for (int j = 1; j <= cnt && i * primes[j] < N; j++)
{
vis[i * primes[j]] = true;
if (i % primes[j] == 0)
{
low[i * primes[j]] = low[i] * j;
if (low[i] == primes[j])
phi[i * primes[j]] = phi[i] * (phi[primes[j]]);
else
phi[i * primes[j]] = phi[i / low[i]] * phi[primes[j] * low[i]];
phi[i * primes[j]] = phi[i] * primes[j];
break;
}
low[i * primes[j]] = primes[j];
phi[i * primes[j]] = phi[i] * phi[primes[j]];
}
}
for (int i = 1; i < N; i++)
s[i] = s[i - 1] + i - phi[i];
}
int main()
{
init();
int T;
scanf("%d", &T);
for (int i = 1, a; i <= T; i++)
{
scanf("%d", &a);
printf("Case %d: %d\n", i, s[a]);
}
return 0;
}
|