牛客练习赛72 D-brz的函数 莫比乌斯反演 + 差分

传送门：https://ac.nowcoder.com/acm/contest/8282/D

题意

$$求解\sum_{i=1}^n\sum_{j=1}^n\mu(ij)$$

思路

$由积性函数的性质，当m和n互质的时候，\mu(mn)=\mu(m)\mu(n)。即$

$$\sum_{i=1}^n\sum_{j=1}^n\mu(ij)=\sum_{i=1}^n\sum_{j=1}^n\mu(i)\mu(j)[gcd(i,j)=1]$$

$$=\sum_{i=1}^n\sum_{j=1}^n\mu(i)\mu(j)\sum_{di\;\;dj}\mu(d)$$

$枚举d：$

$$=\sum_{d=1}^n\mu(d)\sum_{di}\mu(i)\sum_{dj}\mu(j)$$

$$=\sum_{d=1}^n\mu(d)\left ( \sum_{di}^n\mu(i) \right ) ^2$$

$$=\sum_{d=1}^n\mu(d)\left ( \sum_{i=1}^{\frac{n}{d}}\mu(i*d) \right ) ^2$$ 我们枚举d，对于每一个n，都会有$kd\leq n < (k+1)d$，所以在$kd$和$(k+1)d-1$之间都会对答案有贡献，所以可以先差分在取前缀和即可。

Code

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##include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pdd;

##define INF 0x3f3f3f3f
##define lowbit(x) x & (-x)
##define mem(a, b) memset(a , b , sizeof(a))
##define FOR(i, x, n) for(int i = x;i <= n; i++)

// const ll mod = 998244353;
// const ll mod = 1e9 + 7;
// const double eps = 1e-6;
// const double PI = acos(-1);
// const double R = 0.57721566490153286060651209;

const int N = 5e4 + 10;

int mu[N]; // 莫比乌斯函数
bool is_prime[N];
int prime[N];
int cnt;
int ans[N + 10];

void init()
{
    mu[1] = 1;
    is_prime[0] = is_prime[1] = true;
    for(int i = 2;i < N; i++) {
        if (!is_prime[i]) {
            mu[i] = -1;
            prime[++cnt] = i;
        }
        for (int j = 1; j <= cnt && i * prime[j] < N; j++) {
            is_prime[i * prime[j]] = true;
            if (i % prime[j] == 0) {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }

    for(int d = 1;d < N; d++) {
        int tmp = 0;
        for(int i = d;i < N; i += d) {
            int l = i;
            int r = min(l + d - 1, N);
            tmp += mu[l];
            ans[l] += mu[d] * tmp * tmp;
            ans[r + 1] -= mu[d] * tmp * tmp;    
        }
    }

    for(int i = 1;i < N; i++) ans[i] += ans[i - 1];
}

void solve()
{
    init();
    int _;
    cin >> _;
    while(_--) {
        int x;
        cin >> x;
        cout << ans[x] << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    //cin.tie(nullptr);
    //cout.tie(nullptr);
##ifdef FZT_ACM_LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
##else
    solve();
##endif
    return 0;
}

本文作者：jujimeizuo
本文地址： https://blog.jujimeizuo.cn/2020/11/11/nowcoder-practice72-d/
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