AtCoder Beginner Contest 162 C~F

比赛链接：Here

AB水题，

C - Sum of gcd of Tuples (Easy)

题意：$\sum _{a=1}^K\sum _{b=1}^K\sum _{c=1}^{K}gcd(a,b,c)$

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 $S_i\ne S_j,S_i\ne S_k,S_j\ne S_k,j-i\ne k-j_{\circ }$

数据范围：$1\le N\le 4000$

题解：先将满足 ${}_i\ne S_j,S_i\ne S_k,S_j\ne S_k$ 的算出来，在减去 $j-i\ne k-j$ 的数目。

总的显然等于 $num(R)\ast num(B)\ast num(G)$ ，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：$\sum _{a_1=1}^K\sum _{a_2=1}^K\dots \sum _{a_N=1}^{K}gcd(a_1,a_2,\dots ,a_N)(mod1\mathrm{e}9+7)$

数据范围：$2\le N\le {10}^5,1\le K\le {10}^5$

思路一：莫比乌斯反演化简

$$\begin{array}{l}Ans=\sum _{a_1=1}^K\sum _{a_2=1}^K\dots \sum _{a_N=1}^{K}gcd(a_1,a_2,\dots ,a_N)\\ =\sum _{i=1}^K\sum _{a_1=1}^K\sum _{a_2=1}^K\dots \sum _{a_N=1}^{K}i[\gcd (a_1,a_2,\dots ,a_N)==i]\\ =\sum _{i=1}^K\sum _{a_1=1}^{\lfloor \frac{K}{i}\rfloor }\sum _{a_2=1}^{\lfloor \frac{K}{i}\rfloor }\dots \sum _{a_N=1}^{\lfloor \frac{K}{i}\rfloor }i[\gcd (a_1,a_2,\dots ,a_N)==1]\\ =\sum _{i=1}^K\sum _{a_1=1}^{\lfloor \frac{K}{i}\rfloor }\sum _{a_2=1}^{\lfloor \frac{K}{i}\rfloor }\dots \sum _{a_N=1}^{\lfloor \frac{K}{i}\rfloor }i\sum _{d=1}^{\lfloor \frac{K}{i}\rfloor }\mu (d)\lfloor \frac{d}{a_1}\rfloor \lfloor \frac{d}{a_2}\rfloor \dots \lfloor \frac{d}{aN}\rfloor \\ =\sum _{i=1}^{K}i\sum _{d=1}^{\lfloor \frac{K}{ı}\rfloor }\mu (d)\sum _{a_1=1}^{\lfloor \frac{K}{id}\rfloor }\sum _{a2=1}^{\lfloor \frac{K}{id}\rfloor }\dots \sum _{aN=1}^{\lfloor \frac{K}{id}\rfloor }1\\ =\sum _{i=1}^{K}i\sum _{d=1}^{\lfloor \frac{K}{i}\rfloor }\mu (d){\lfloor \frac{K}{id}\rfloor }^N\\ =\sum _{T=1}^K{\lfloor \frac{K}{T}\rfloor }^N\sum _{d\mid T}\mu (d)\ast \lfloor \frac{T}{d}\rfloor (T=id)\\ =\sum _{T=1}^K{\lfloor \frac{K}{T}\rfloor }^N\varphi (T)\end{array}$$

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：$f[i]={\lfloor \frac{K}{i}\rfloor }^N-\sum _{j>i,i\mid j}f[j{]}_{\circ }$​

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 $[\frac{N}{2}\rfloor$ 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， $B_1,B_2\dots ,B_{\lfloor \frac{N}{2}\rfloor }$，可以发现 $\sum _{i=2}^{\lfloor \frac{N}{2}\rfloor }{B}_i-B_{i-1}-2\le 2$

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

$$\{\begin{array}{rl}f[i][0]& =f[i-2][0]+a[i]\\ f[i][1]& =max(f[i-2][1],f[i-3][0])+a[i]\\ f[i][2]& =max(f[i-2][2],f[i-3][1],f[i-4][0])+a[i]\end{array}$$

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;
}