I. Rise of Shadows

一天有$H$個小時，$M$分鐘，問，有多少個整數分鐘，滿足時針與分針的角度$\le \alpha$，$\alpha =\frac{2\pi A}{HM}$。
$$\begin{gathered} \sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[\mid \frac{2\pi (i\times M+j)}{HM}?\frac{2\pi j}{M}\mid \le \frac{2\pi A}{HM}] \\ \sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[\mid i\times M+j?H\times j\mid \le A] \\ H\times M?\sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i\times M+j?H\times j>A]?\sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i\times M+j?H\times j<?A] \end{gathered}$$

$$\begin{gathered} \sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i\times M+j?H\times j>A] \\ \sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i>\frac{(H?1)\times j+A}{M}] \\ \sum _{j=0}^{M?1}\left(H?1?\frac{(H?1)\times j+A}{M}\right) \\ M\times (H?1)?\sum _{i=0}^{M?1}\frac{(H?1)\times i+A}{M} \end{gathered}$$

$$\begin{gathered} \sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i\times M+j?H\times j<?A] \\ \sum _{i=0}^{H?1}\left(M?\sum _{j=0}^{M?1}[i\times M+j?H\times j\ge ?A]\right) \\ H\times M?\sum _{i=0}^{H?1}\sum _{j=0}^{M?1}[i\times M+j?H\times j>?A?1+(H+1)?(H+1)] \\ H\times M?\sum _{j=0}^{M?1}\sum _{i=0}^{H?1}[i>\frac{(H?1)\times j+(H+1)M?A?1)}{M}?(H+1)] \\ H\times M?\sum _{j=0}^{M?1}H?1?\frac{(H?1)\times j+(H+1)M?A?1}{M}+H+1 \\ ?HM+\sum _{i=0}^{M?1}\frac{(H?1)\times i+(H+1)M?A?1}{M} \end{gathered}$$
綜上，答案為：
$$M+HM+\sum _{i=0}^{M?1}\frac{(H?1)\times i+A}{M}?\sum _{i=0}^{M?1}\frac{(H?1)\times i+(H+1)M?A+1}{M}$$

#include <bits/stdc++.h> #define int long longusing namespace std;long long f(long long a, long long b, long long c, long long n) {if (!a) {return (b / c) * (n + 1);}if (a >= c || b >= c) {return f(a % c, b % c, c, n) + (b / c) * (n + 1) + (a / c) * n * (n + 1) / 2;}long long m = (a * n + b) / c;return n * m - f(c, c - b - 1, a, m - 1); }signed main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);int h, m, a;cin >> h >> m >> a;cout << min(m + h * m + f(h - 1, a, m, m - 1) - f(h - 1, (h + 1) * m - a - 1, m, m - 1), h * m) << "\n";return 0; }

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