The Center of Gravity Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3971 Accepted Submission(s): 2280 Problem Description Everyone know the story that how Newton discovered the Universal Gr
public static void main(String[] args) throws Exception { String queryForScanUsers_SQL = "select a.username AS user_name,b.* from tbl_aaa a left join tbl_bbb b where a.id=? limit 10"; //去除WHERE和LIMIT String sqlRegex = "(select\\s+)(.*)(\\s+
Python之进度条及π的计算 文本进度条 1. 简单的开始 这是利用print()函数来实现简单的非刷新文本进度条.它的基本思想是按照任务执行百分比将整个任务划分为100个单位,每执行N%输出一次进度条. 为了模拟任务处理的时间效果,需要调用Python标准时间库time. 完整代码如下: import time scale=10 print("-----执行开始-----") for i in range(scale+1): a,b='**'*i,'..'*(scale-i) c
python求100以内素数之和 from math import sqrt # 使用isPrime函数 def isPrime(n): if n <= 1: return False for i in range(2, int(sqrt(n)) + 1): if n % i == 0: return False return True count = 0 for i in range(101): if isPrime(i): count += i print(count) # 单行程序扫描素数
Line.h #pragma once //Microsoft Visual Studio 2015 Enterprise //根据两点式方法求直线,并求两条直线的交点 #include"BoundaryPoint.h" #include"Coordinates.h" class Line { public: Line GetLine(BoundaryPoint sourcePoint, BoundaryPoint endPoint); Line GetLine(C
You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13549 Accepted Submission(s): 6645 Problem Description Many geometry(几何)problems were designed in the ACM/