luogu1447 [NOI2010]能量采集 莫比乌斯反演

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冬令营考炸了，我这个菜鸡只好颓废数学题了

NOI2010能量采集

由题意可以写出式子：

(sum_{i=1}^nsum_{j=1}^m(2gcd(i,j)-1))

(=2sum_{i=1}^nsum_{j=1}^mgcd(i,j)-nm)

我们现在考虑(sum_{i=1}^nsum_{j=1}^mgcd(i,j))，默认n比m小

(=sum_{p=1}^npsum_{i=1}^nsum_{j=1}^m[gcd(i,j)=p])

(=sum_{p=1}^npsum_{i=1}^{n/p}sum_{j=1}^{m/p}[gcd(i,j)=1])

(=sum_{p=1}^npsum_{i=1}^{n/p}sum_{j=1}^{m/p}sum_{d|i,d|j}mu(d))

(=sum_{p=1}^npsum_{d=1}^nmu(d)lfloorfrac n{pd} floorlfloorfrac m{pd} floor)

(=sum_{q=1}^nsum_{d|q}mu(d)frac qdlfloorfrac n{q} floorlfloorfrac m{q} floor)

由于是单组数据，所以不用前缀和数论分块

所以这是一道莫比乌斯反演题，32行一遍AC

#include <cstdio>
using namespace std;

bool vis[100010];
int mu[100010], tot, prime[100010], fuck = 100000;
long long sum[100010];

int main()
{
	mu[1] = 1;
	for (int i = 2; i <= fuck; i++)
	{
		if (vis[i] == 0) prime[++tot] = i, mu[i] = -1;
		for (int j = 1; j <= tot && i * prime[j] <= fuck; j++)
		{
			vis[i * prime[j]] = true;
			if (i % prime[j] == 0) break;
			mu[i * prime[j]] = -mu[i];
		}
	}
	for (int i = 1; i <= fuck; i++)
		for (int j = i, k = 1; j <= fuck; j += i, k++)
			sum[j] += mu[i] * k;
	int n, m;
	scanf("%d%d", &n, &m);
	if (n > m) { int t = n; n = m; m = t; }
	long long ans = 0;
	for (int i = 1; i <= n; i++)
		ans += sum[i] * (n / i) * (m / i);
	printf("%lld
", ans * 2 - n * (long long)m);
	return 0;
}