做网站什么主题比较好,做商城网站如何寻找货源,查公司查企业用什么软件,有哪些平面设计网站I. Rise of Shadows
一天有HHH个小时#xff0c;MMM分钟#xff0c;问#xff0c;有多少个整数分钟#xff0c;满足时针与分针的角度≤α\le \alpha≤α#xff0c;α2πAHM\alpha \frac{2 \pi A}{HM}αHM2πA。 ∑i0H−1∑j0M−1[∣2π(iMj)HM−2πjM∣≤2πAHM]∑i0…I. Rise of Shadows
一天有HHH个小时MMM分钟问有多少个整数分钟满足时针与分针的角度≤α\le \alpha≤αα2πAHM\alpha \frac{2 \pi A}{HM}αHM2πA。 ∑i0H−1∑j0M−1[∣2π(i×Mj)HM−2πjM∣≤2πAHM]∑i0H−1∑j0M−1[∣i×Mj−H×j∣≤A]H×M−∑i0H−1∑j0M−1[i×Mj−H×jA]−∑i0H−1∑j0M−1[i×Mj−H×j−A]\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]\\ i0∑H−1j0∑M−1[∣HM2π(i×Mj)−M2πj ∣≤HM2πA]i0∑H−1j0∑M−1[∣i×Mj−H×j∣≤A]H×M−i0∑H−1j0∑M−1[i×Mj−H×jA]−i0∑H−1j0∑M−1[i×Mj−H×j−A]
∑i0H−1∑j0M−1[i×Mj−H×jA]∑i0H−1∑j0M−1[i(H−1)×jAM]∑j0M−1(H−1−(H−1)×jAM)M×(H−1)−∑i0M−1(H−1)×iAM\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}\\ i0∑H−1j0∑M−1[i×Mj−H×jA]i0∑H−1j0∑M−1[iM(H−1)×jA]j0∑M−1(H−1−M(H−1)×jA)M×(H−1)−i0∑M−1M(H−1)×iA
∑i0H−1∑j0M−1[i×Mj−H×j−A]∑i0H−1(M−∑j0M−1[i×Mj−H×j≥−A])H×M−∑i0H−1∑j0M−1[i×Mj−H×j−A−1(H1)−(H1)]H×M−∑j0M−1∑i0H−1[i(H−1)×j(H1)M−A−1)M−(H1)]H×M−∑j0M−1H−1−(H−1)×j(H1)M−A−1MH1−HM∑i0M−1(H−1)×i(H1)M−A−1M\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}\\ i0∑H−1j0∑M−1[i×Mj−H×j−A]i0∑H−1(M−j0∑M−1[i×Mj−H×j≥−A])H×M−i0∑H−1j0∑M−1[i×Mj−H×j−A−1(H1)−(H1)]H×M−j0∑M−1i0∑H−1[iM(H−1)×j(H1)M−A−1)−(H1)]H×M−j0∑M−1H−1−M(H−1)×j(H1)M−A−1H1−HMi0∑M−1M(H−1)×i(H1)M−A−1 综上答案为 MHM∑i0M−1(H−1)×iAM−∑i0M−1(H−1)×i(H1)M−A1MM 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} MHMi0∑M−1M(H−1)×iA−i0∑M−1M(H−1)×i(H1)M−A1
#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;
}
