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二重积分参考代码
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class doubleIntegral: def __init__(self, xMin, xMax, yMin, yMax, func): self.xMin = xMin self.xMax = xMax self.yMin = yMin self.yMax = yMax self.func = func def trapezoid(self, nX, nY): result = 0 betweenX = (self.xMax - self.xMin) / nX betweenY = (self.yMax - self.yMin) / nY sums = [] for i in range(nY + 1): temp = 0 for j in range(nX): temp += betweenX / 2 * (self.func(self.xMin + j * betweenX, self.yMin + i * betweenY) + self.func(self.xMin + (j + 1) * betweenX, self.yMin + i * betweenY)) sums.append(temp) ans = 0 for i in range(nY): ans += betweenY / 2 * (sums[i] + sums[i + 1]) return ans def simpsons(self, nX, nY): result = 0 betweenX = (self.xMax - self.xMin) / nX betweenY = (self.yMax - self.yMin) / nY / 2 sums = [] for i in range(2 * nY + 1): temp = 0 for j in range(nX): a = self.xMin + j * betweenX b = self.xMin + (j + 1) * betweenX c = (a + b) / 2 temp += betweenX / 6 * (self.func(a, self.yMin + i * betweenY) + self.func( b, self.yMin + i * betweenY) + 4 * self.func(c, self.yMin + i * betweenY)) sums.append(temp) ans = 0 for i in range(nY): ans += betweenY / 3 * \ (sums[i * 2] + sums[(i + 1) * 2] + 4 * sums[i * 2 + 1]) return ans if __name__ == "__main__": # 个人不知道为什么梯形法求出来不对，有兴趣的可以自己去看看 # 输入格式：xMin, xMax, yMin, yMax, func # 输入格式'： nx, ny print(doubleIntegral(0, 1, 0, 2, lambda x, y: x**2+2*y).trapezoid(4, 8)) print(doubleIntegral(0, 1, 0, 2, lambda x, y: x**2+2*y).simpsons(4, 8)) |
树莓派4b jupyter frp端口映射
树莓派 frpc 设置：
从 github 中找到 arm 版本 wget 下载
frpc.ini 参考配置
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[common] server_addr = you server ip server_port = 7000 token = your password [jupyter] type = http local_ip = 127.0.0.1 local_port = 8888 custom_domains = your domain |
jupyter 设置跟之前文章相同，即 ip, allow_remote_access等设置
服务器端 frps 配置: 云端开启7000~7003端口
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[common] bind_port = 7000 vhost_http_port = 7002 token = your password dashboard_port = 7001 dashboard_user = Your dashboard usr name dashboard_pwd = Your dashboard password vhost_http_port = 7003 |
云端 nginx 参考配置：
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server{ listen 80; listen [::]:80; server_name yourdomain; location / { proxy_pass http://127.0.0.1:7003; proxy_set_header X-Real-IP $remote_addr; proxy_set_header Host $host; proxy_set_header X-Forwarded-For $proxy_add_x_forwarded_for; proxy_http_version 1.1; proxy_set_header Upgrade $http_upgrade; proxy_set_header Connection "upgrade"; proxy_read_timeout 120s; proxy_next_upstream error; } } |
最终效果即通过域名 python.pancake2021.work 访问树莓派 jupyter
计算数学 ODE LAB 3
全部是正确答案，慎用。Adams和Milne需要用另外的算法如runge-kutta（代码示例求法）求前3个数，当然也可以用Euler和Euler Improved，大家不要都一样啦。Python缩进规范请注意！
Python Euler Improved
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def solveByEulerImproved(f, epsilon, a, y_a, b): function = Result.get_function(f) n = 1 delta = float("inf") while abs(delta) >= epsilon / 10: n *= 2 h = (b - a) / n pre_y = y_a pre_fy = function(a, y_a) for i in range(n): x = a + (i + 1 / 2) * h yHalf = pre_y + pre_fy * h / 2 fyHalf = function(x, yHalf) y = pre_y + h * fyHalf delta = pre_y - y pre_y = y pre_fy = function(x + h / 2, y) return y |
Python Euler
Euler method 理论上代码正确，但是实际上因为算法的原因，与测试样例的值偏差会过大，要通过测试样例只需要初始化时把 n = 10000000
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def solveByEuler(f, epsilon, a, y_a, b): function = Result.get_function(f) n = 1 delta = float("inf") while abs(delta) >= epsilon / 10: n *= 2 h = (b - a) / n pre_y = y_a for i in range(n): x = a + i * h fy = function(x, pre_y) y = pre_y + h * fy delta = pre_y - y pre_y = y return y |
Python Runge-Kutta
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def solveByRungeKutta(f, epsilon, a, y_a, b): function = Result.get_function(f) n = 1 delta = float("inf") while abs(delta) >= epsilon / 10: n *= 2 h = (b - a) / n pre_y = y_a for i in range(n): x = a + i * h k1 = h * function(x, pre_y) k2 = h * function(x + h / 2, pre_y + k1 / 2) k3 = h * function(x + h / 2, pre_y + k2 / 2) k4 = h * function(x + h, pre_y + k3) y = (k1 + 2 * k2 + 2 * k3 + k4) / 6 + pre_y delta = pre_y - y pre_y = y return y |
Python Adams
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def runge_kutta(h, a, y_a, function): y_dot = [function(a, y_a)] pre_y = y_a for i in range(3): x = a + i * h k1 = h * function(x, pre_y) k2 = h * function(x + h / 2, pre_y + k1 / 2) k3 = h * function(x + h / 2, pre_y + k2 / 2) k4 = h * function(x + h, pre_y + k3) y = (k1 + 2 * k2 + 2 * k3 + k4) / 6 + pre_y y_dot.append(function(x + h, y)) pre_y = y return y, y_dot def solveByAdams(f, epsilon, a, y_a, b): function = Result.get_function(f) n = 2 delta = float("inf") while abs(delta) >= epsilon / 10: n *= 2 h = (b - a) / n y, y_dot = Result.runge_kutta(h, a, y_a, function) for i in range(n - 3): delta = (55 * y_dot[-1] - 59 * y_dot[-2] + 37 * y_dot[-3] - 9 * y_dot[-4]) / 24 * h y += delta y_dot.append(function(a + (i + 4) * h, y)) return y |
Python Milne
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def runge_kutta(h, a, y_a, function): y_dot = [function(a, y_a)] y_ = [y_a] pre_y = y_a for i in range(3): x = a + i * h k1 = h * function(x, pre_y) k2 = h * function(x + h / 2, pre_y + k1 / 2) k3 = h * function(x + h / 2, pre_y + k2 / 2) k4 = h * function(x + h, pre_y + k3) y = (k1 + 2 * k2 + 2 * k3 + k4) / 6 + pre_y y_dot.append(function(x + h, y)) y_.append(y) pre_y = y return y_, y_dot def solveByMilne(f, epsilon, a, y_a, b): function = Result.get_function(f) n = 2 delta = float("inf") while abs(delta) >= epsilon / 10: n *= 2 h = (b - a) / n y, y_dot = Result.runge_kutta(h, a, y_a, function) for i in range(n - 3): y.append(y[-4] + 4 * h / 3 * (2 * y_dot[-1] - y_dot[-2] + 2 * y_dot[-3])) y_dot.append(function(a + (i + 4) * h, y[-1])) y[-1] = y[-3] + h / 3 * (y_dot[-1] + 4 * y_dot[-2] + y_dot[-3]) delta = y[-1] - y[-2] y_dot[-1] = function(a + (i + 4) * h, y[-1]) return y[-1] |
计算数学 ODE Python参考代码
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class ODE: def __init__(self, function, x0, y0): self.function = function self.x0 = x0 self.y0 = y0 def euler(self, h, step): result = [] pre_y = self.y0 for i in range(step): x = self.x0 + i * h fy = self.function(x, pre_y) y = pre_y + h * fy result.append(y) pre_y = y # print("y_{} = {}".format(i + 1, y)) return result def eulerImproved(self, h, step): result = [] pre_y = self.y0 pre_fy = self.y0 for i in range(step): x = self.x0 + (i + 1 / 2) * h yHalf = pre_y + pre_fy * h / 2 fyHalf = self.function(x, yHalf) y = pre_y + h * fyHalf result.append(y) pre_y = y pre_fy = function(x + h / 2, y) # print("y_{} = {}".format(i + 1, y)) return result def runge_kutta(self, h, step): result = [] pre_y = self.y0 for i in range(step): x = self.x0 + i * h k1 = h * self.function(x, pre_y) k2 = h * self.function(x + h / 2, pre_y + k1 / 2) k3 = h * self.function(x + h / 2, pre_y + k2 / 2) k4 = h * self.function(x + h, pre_y + k3) y = (k1 + 2 * k2 + 2 * k3 + k4) / 6 + pre_y result.append(y) pre_y = y # print("y_{} = {}".format(i + 1, y)) return result def milne(self, h, step): y = [self.y0] + self.runge_kutta(h, 3) y_dot = [function(self.x0 + i * h, y[i]) for i in range(4)] for i in range(step - 3): y.append(y[-4] + 4 * h / 3 * (2 * y_dot[-1] - y_dot[-2] + 2 * y_dot[-3])) y_dot.append(function(self.x0 + (i + 4) * h, y[-1])) y[-1] = y[-3] + h / 3 * (y_dot[-1] + 4 * y_dot[-2] + y_dot[-3]) y_dot[-1] = function(self.x0 + (i + 4) * h, y[-1]) # print("y_{} = {}".format(i + 4, y[-1])) return y[1:] def adams(self, h, step): y = [self.y0] + self.runge_kutta(h, 3) y_dot = [function(self.x0 + i * h, y[i]) for i in range(4)] for i in range(step - 3): y.append(y[-1] + (55 * y_dot[-1] - 59 * y_dot[-2] + 37 * y_dot[-3] - 9 * y_dot[-4]) / 24 * h) y_dot.append(function(self.x0 + (i + 4) * h, y[-1])) # print("y_{} = {}".format(i + 4, y[-1])) return y[1:] if __name__ == "__main__": function = lambda x, y: y - 2 * x / y x0 = 0 y0 = 1 h = 0.002 step = 1000 ode = ODE(function, x0, y0) print(ode.euler(h, step)[-1]) print("--------------------------------------------") print(ode.eulerImproved(h, step)[-1]) print("--------------------------------------------") print(ode.runge_kutta(h, step)[-1]) print("--------------------------------------------") print(ode.adams(h, step)[-1]) print("--------------------------------------------") print(ode.milne(h, step)[-1]) |
Timus #1162. Currency Exchange
题目：一共N个货币种类，当前为S种类的货币，当前拥有的该货币价值为V，下面M行包含可以交换货币的汇率/手续费信息：A,B,RAB,CAB,RBA,CBA分别代表种类A，种类B，A到B的汇率，A到B的手续费，B到A的汇率，B到A的手续费。求能否通过某种交换方式能让货币越来越多即大于V。
解析：Bellman-Ford。修改#1450代码即可
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#include<iostream> #include<vector> #include<queue> #define INF 0x3f3f3f3f using namespace std; int n, m, s; float v; struct Currency { Currency() { money = INF; done = false; } //neighbor_index rate commission vector<pair<int, pair<float, float>>> neighbor; float money; bool done; }currency[100]; void SPFA() { queue<int> q; currency[s - 1].money = -v; q.push(s - 1); while (!q.empty()) { int x = q.front(); q.pop(); currency[x].done = false; for (auto ptr = currency[x].neighbor.begin(); ptr != currency[x].neighbor.end(); ptr++) { if (currency[ptr->first].money > -(-currency[x].money - ptr->second.second) * ptr->second.first) { currency[ptr->first].money = -(-currency[x].money - ptr->second.second) * ptr->second.first; if (!currency[ptr->first].done) { currency[ptr->first].done = true; q.push(ptr->first); } } } } if (currency[s - 1].money < v)return; } int main() { //货币种类 下面共几行 当前货币种类 当前拥有的该货币价值 cin >> n >> m >> s >> v; for (int i = 0; i < m; i++) { int a, b; float rab, cab, rba, cba; cin >> a >> b >> rab >> cab >> rba >> cba; currency[a - 1].neighbor.push_back(make_pair(b - 1, make_pair(rab, cab))); currency[b - 1].neighbor.push_back(make_pair(a - 1, make_pair(rba, cba))); } SPFA(); if (-currency[s - 1].money <= v) cout << "NO" << endl; else cout << "YES" << endl; } |
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