「NOI2016」循环之美

传送门

这个题面一眼看过去没什么想法，尝试转化题意。

假设我们现在有合法的一组 (x, y)，显然我们先要有 (x perp y)。

设 (frac{x}{y}) 在 (k) 进制下是一个循环节长度为 (l) 的循环小数。

那么我们显然可以得到 ([frac{xk^l}{y}] = [frac{x}{y}])，也就是两者的小数部分相等。

那么我们可以进一步推导：

[egin{split} [frac{xk^l}{y}] = [frac{x}{y}] & Rightarrow frac{xk^l}{y} - lfloor frac{xk^l}{y} floor = frac{x}{y} - lfloor frac{x}{y} floor \ & Rightarrow xk^l - lfloor frac{xk^l}{y} floor imes y = x - lfloor frac{x}{y} floor imes y \ & Rightarrow xk^l equiv x (mod y) \ & Rightarrow k^l equiv 1 (mod y)quadquad(gcd(x, y) = 1) \ & Rightarrow k perp y end{split} ]

推导到这里我们就不难发现其实我们是要求一个这样的东西：

[f(n, m, k) = sum_{i = 1} ^ n sum_{j = 1} ^ m [i perp j] [j perp k] ]

考虑化简后面那一坨：

[egin{split} f(n, m, k) & = sum_{i = 1} ^ n sum_{j = 1} ^ m [i perp j] [j perp k] \ & = sum_{i = 1} ^ n sum_{j = 1} ^ m [i perp j] sum_{d | j, d | k} mu(d) \ & = sum_{d = 1} ^ {min(n, m)} mu(d) sum_{d | j} sum_{i = 1} ^ n [i perp j] \ & = sum_{d = 1} ^ {min(n, m)} mu(d) sum_{j = 1} ^ {lfloor frac{m}{d} floor} sum_{i = 1} ^ n [j perp i] [i perp d] \ & = sum_{d = 1} ^ {min(n, m)} mu(d) f(lfloor frac{m}{d} floor, n, d) end{split} ]

边界：

(f(n, 0, k) = f(0, m, k) = 0, f(n, m, 1) = sum_{i = 1} ^ n sum_{j = 1} ^ m [i perp j] = sum_{d = 1} ^ {min(n, m)} mu(d) lfloor frac{n}{d} floor lfloor frac{m}{d} floor)

杜教筛 (mu)，然后递归求就是了。

参考代码：

#include <cstring>
#include <cstdio>
#include <vector>
#include <map>
using namespace std;

typedef long long LL;
const int _ = 1e7 + 5;

int vis[_], num, pri[_]; LL mu[_], s[_];
vector < int > fac;
map < int, LL > ans;

void seive() {
    mu[1] = 1;
    for (int i = 2; i < _; ++i) {
        if (!vis[i]) mu[i] = -1, pri[++num] = i;
        for (int j = 1; j <= num && i * pri[j] < _; ++j) {
            vis[i * pri[j]] = 1;
            if (i % pri[j] == 0) break ;
            mu[i * pri[j]] = -mu[i];
        }
    }
    for (int i = 1; i < _; ++i) s[i] = s[i - 1] + mu[i];
}

LL Sum(int n) {
    if (n < _) return s[n]; else if (ans[n]) return ans[n];
    LL res = 1;
    for (int l = 2, r; l <= n; l = r + 1)
        r = n / (n / l), res -= 1ll * (r - l + 1) * Sum(n / l);
    return ans[n] = res;
}

LL solve(int n, int m, int k) {
    if (!n || !m) return 0;
    LL res = 0;
    if (k == 1) {
        for (int l = 1, r; l <= min(n, m); l = r + 1)
            r = min(n / (n / l), m / (m / l)), res += 1ll * (Sum(r) - Sum(l - 1)) * (n / l) * (m / l);
        return res;
    }
    for (int i = 0; i < fac.size(); ++i)
        if (k % fac[i] == 0 && mu[fac[i]])
            res += mu[fac[i]] * solve(m / fac[i], n, fac[i]);
    return res;
}

int n, m, k;

int main() {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin), freopen("cpp.out", "w", stdout);
#endif
    seive();
    scanf("%d %d %d", &n, &m, &k);
    for (int i = 1; i * i <= k; ++i)
        if (k % i == 0) {
            fac.push_back(i);
            if (i * i != k) fac.push_back(k / i);
        }
    printf("%lld
", solve(n, m, k));
    return 0;
}