AtCoder Beginner Contest 162 C~F

比赛链接：Here

AB水题，

C - Sum of gcd of Tuples (Easy)

题意：\(\sum_{a=1}^{K} \sum_{b=1}^{K} \sum_{c=1}^{K} g c d(a, b, c)\)

数据范围：\(1\le K\le 200\)

思路：因为 \(k\) 比较小，我们直接跑暴力即可

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n; cin >> n;
    ll ans = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= n; k++)
                ans += __gcd(__gcd(i, j), k);
    cout << ans;
}

D - RGB Triplets

题意：给一个长度为N的字符串S，只包含'R'，'B'，'G'。求有多少个三元组 \((i,j,k)(1\le i<j<k\le N)\) 满足 \(S_{i} \neq S_{j}, S_{i} \neq S_{k}, S_{j} \neq S_{k}, j-i \neq k-j_{\circ}\)

数据范围：\(1\le N \le 4000\)

题解：先将满足 \(_{i} \neq S_{j}, S_{i} \neq S_{k}, S_{j} \neq S_{k}\) 的算出来，在减去 \(j-i \neq k-j\) 的数目。

总的显然等于 \(num(R)*num(B)*num(G)\) ，然后枚举两个端点，判断第三个端点是不是不同于这个两个端点的颜色。

const int N = 4e3 + 5;
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n; string s;
    cin >> n >> s;
    int a = 0, b = 0, c = 0;
    for (int i = 0; i < n; ++i) {
        if (s[i] == 'R') a += 1;
        if (s[i] == 'G') b += 1;
        if (s[i] == 'B') c += 1;
    }
    ll ans = 1ll * a * b * c;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++) {
            if (s[i] == s[j]) continue;
            if (2 * j - i < n && s[i] != s[2 * j - i] && s[j] != s[2 * j - i]) ans--;
        }
    cout << ans;
}

E - Sum of gcd of Tuples (Hard)

题意：\(\sum_{a_{1}=1}^{K} \sum_{a_{2}=1}^{K} \ldots \sum_{a_{N}=1}^{K} g c d\left(a_{1}, a_{2}, \ldots, a_{N}\right)(\bmod 1 \mathrm{e} 9+7)\)

数据范围：\(2 \leq N \leq 10^{5}, 1 \leq K \leq 10^{5}\)

思路一：莫比乌斯反演化简

\[\begin{array}{l}
A n s=\sum_{a_{1}=1}^{K} \sum_{a_{2}=1}^{K} \ldots \sum_{a_{N}=1}^{K} g c d\left(a_{1}, a_{2}, \ldots, a_{N}\right) \\
=\sum_{i=1}^{K} \sum_{a_{1}=1}^{K} \sum_{a_{2}=1}^{K} \ldots \sum_{a_{N}=1}^{K} i\left[\operatorname{gcd}\left(a_{1}, a_{2}, \ldots, a_{N}\right)==i\right] \\
=\sum_{i=1}^{K} \sum_{a_{1}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \sum_{a_{2}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \ldots \sum_{a_{N}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} i\left[\operatorname{gcd}\left(a_{1}, a_{2}, \ldots, a_{N}\right)==1\right] \\
=\sum_{i=1}^{K} \sum_{a_{1}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \sum_{a_{2}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \ldots \sum_{a_{N}=1}^{\left\lfloor\frac{K}{i}\right\rfloor} i \sum_{d=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \mu(d)\left\lfloor\frac{d}{a_{1}}\right\rfloor\left\lfloor\frac{d}{a_{2}}\right\rfloor \ldots\left\lfloor\frac{d}{a N}\right\rfloor \\
=\sum_{i=1}^{K} i \sum_{d=1}^{\left\lfloor\frac{K}{\imath}\right\rfloor} \mu(d) \sum_{a_{1}=1}^{\left\lfloor\frac{K}{i d}\right\rfloor} \sum_{a 2=1}^{\left\lfloor\frac{K}{i d}\right\rfloor} \ldots \sum_{a N=1}^{\left\lfloor\frac{K}{i d}\right\rfloor} 1 \\
=\sum_{i=1}^{K} i \sum_{d=1}^{\left\lfloor\frac{K}{i}\right\rfloor} \mu(d)\left\lfloor\frac{K}{i d}\right\rfloor^{N} \\
=\sum_{T=1}^{K}\left\lfloor\frac{K}{T}\right\rfloor^{N} \sum_{d \mid T} \mu(d) *\left\lfloor\frac{T}{d}\right\rfloor(T=i d) \\
=\sum_{T=1}^{K}\left\lfloor\frac{K}{T}\right\rfloor^{N} \phi(T)
\end{array}
\]

由于 \(K\) 不大，预处理欧拉函数，直接遍历即可。\(K\) 大的话，整除分块加杜教筛（雾。

#include <bits/stdc++.h>
using ll = long long;
using namespace std;
const int N = 1e5 + 5, MD = 1e9 + 7;
int pri[N], tot, phi[N];
bool p[N];
void init() {
    p[1] = true, phi[1] = 1;
    for (int i = 2; i < N; i++) {
        if (!p[i]) pri[++tot] = i, phi[i] = i - 1;
        for (int j = 1; j <= tot && i * pri[j] < N; j++) {
            p[i * pri[j]] = true;
            if (i % pri[j] == 0) {
                phi[i * pri[j]] = phi[i] * pri[j];
                break;
            } else phi[i * pri[j]] = phi[i] * (pri[j] - 1);
        }
    }
}
ll qpow(ll a, ll b) {
    ll ans = 1;
    for (; b; b >>= 1, a = a * a % MD) if (b & 1) ans = ans * a % MD;
    return ans;
}
void add(int &x, int y) { x += y; if (x >= MD) x -= MD;}
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    init();
    int n, k, ans = 0;
    cin >> n >> k;
    for (int i = 1; i <= k; i++)
        add(ans, 1LL * qpow(k / i, n)*phi[i] % MD);

    cout << ans;
}

思路二：

学习了下官方题解：定义 \(f[i]\) 代表 \(\gcd\) 为 \(i\) 的个数，递推关系式：\(f[i]=\left\lfloor\frac{K}{i}\right\rfloor^{N}-\sum_{j>i, i \mid j} f[j]_{\circ}\)​

双重循环即可，里面那层循环的复杂度总和是个调和级数，\(\log\) 级别的。

const int N = 1e5 + 5, MD = 1e9 + 7;
int f[N];
ll qpow(ll a, ll b) {
    ll ans = 1;
    for (; b; b >>= 1, a = a * a % MD) if (b & 1) ans = ans * a % MD;
    return ans;
}
void add(int &x, int y) {
    x += y;
    if (x >= MD) x -= MD;
    if (x < 0) x += MD;
}
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, k;
    scanf("%d%d", &n, &k);
    cin >> n >> k;
    for (int i = k; i >= 1; i--) {
        f[i] = qpow(k / i, n);
        for (int j = 2 * i; j <= k; j += i)
            add(f[i], -f[j]);
    }
    int ans = 0;
    for (int i = 1; i <= k; i++)
        add(ans, 1LL * f[i]*i % MD);

    cout << ans;
}

F - Select Half

题意：给一个长度为 \(N\) 的序列 \(A\)，要求选 \(\left[\frac{N}{2}\right\rfloor\) 个数，且下标互不相邻，最大化它们的总和。

数据范围：\(2\le N\le 2\times 10^5\)

题解：对于选的数的下标， \(B_{1}, B_{2} \ldots, B_{\left\lfloor\frac{N}{2}\right\rfloor}\)，可以发现 \(\sum_{i=2}^{\left\lfloor\frac{N}{2}\right\rfloor} B_{i}-B_{i-1}-2 \leq 2\)

因此定义 \(f[i][j]\)​ 代表选第 \(1~i\) 里面的数（必选第个 \(i\) 数），前面多空了 \(j\) 的最大值。

\[\left\{\begin{aligned}
f[i][0] &=f[i-2][0]+a[i] \\
f[i][1] &=\max (f[i-2][1], f[i-3][0])+a[i] \\
f[i][2] &=\max (f[i-2][2], f[i-3][1], f[i-4][0])+a[i]
\end{aligned}\right.
\]

const int N = 2e5 + 5;
int a[N];
ll f[N][3];
int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n; cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; i++)
        for (int j = 0; j < 3; j++)
            f[i][j] = -1e18;

    for (int i = 1; i <= 3; i++)
        f[i][i - 1] = a[i];

    f[3][0] = a[1] + a[3];
    for (int i = 4; i <= n; i++) {
        f[i][0] = f[i - 2][0] + a[i];
        f[i][1] = max(f[i - 2][1], f[i - 3][0]) + a[i];
        f[i][2] = max(f[i - 2][2], f[i - 3][1]) + a[i];
        if (i > 4) f[i][2] = max(f[i][2], f[i - 4][0] + a[i]);
    }
    ll ans;
    if (n & 1) ans = max(f[n][2], max(f[n - 1][1], f[n - 2][0]));
    else
        ans = max(f[n][1], f[n - 1][0]);
    cout << ans;
}