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：

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

\(=2\sum_{i=1}^n\sum_{j=1}^m\gcd(i,j)-nm\)

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

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

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

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

\(=\sum_{p=1}^np\sum_{d=1}^n\mu(d)\lfloor\frac n{pd}\rfloor\lfloor\frac m{pd}\rfloor\)

\(=\sum_{q=1}^n\sum_{d|q}\mu(d)\frac qd\lfloor\frac n{q}\rfloor\lfloor\frac m{q}\rfloor\)

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\n", ans * 2 - n * (long long)m);
    return 0;
}

标签：lfloor,luogu1447,frac,gcd,int,sum,mu,反演,NOI2010
来源： https://www.cnblogs.com/oier/p/10334832.html