最近工作中的一个业务需要再确定范围内取出一个随机数,网上到时搜出来一堆可以实现的方法,但是鄙人仍是理解不了,虽说已经copy方法直接完成工作了.今天抽时间整理下,希望能够帮助大家更好的理解随机数的实现原理: 1.网上直接找到的实现方法: 第一种实现方法: public static String getRandom(int min, int max){ Random random = new Random(); int s = random.nextInt(max) % (max - min +
1.random.nextInt(num) public static void main(String args[]) { Random rdom = new Random(); int max = 1314; int min = 520; for(int i = 0;i<2000;i++){ int count = rdom.nextInt((max - min + 1)) + min; if(count==1313){ System.out.println("1313========
Minimum Sum Time Limit: 16000/8000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3769 Accepted Submission(s): 872 Problem Description You are given N positive integers, denoted as x0, x1 ... xN-1. Then give you
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5101 Accepted Submission(s): 2339 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping abilit
No Pain No Game Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2000 Accepted Submission(s): 851 Problem Description Life is a game,and you lose it,so you suicide. But you can not kill yours
Group Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1959 Accepted Submission(s): 1006 Problem Description There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i
java保留两位小数问题: 方式一: 四舍五入 double f = 111231.5585; BigDecimal b = new BigDecimal(f); double f1 = b.setScale(2, BigDecimal.ROUND_HALF_UP).doubleValue(); 保留两位小数 --------------------------------------------------------------- 方
R.java 文件内报错:Underscores can only be used with source level 1.7 or greater 网上查找后得知是Android工程图片资源命名的问题,具体参考http://black-tulip.blogcn.com/2012/08/android%E5%B7%A5%E7%A8%8B%E5%9B%BE%E7%89%87%E8%B5%84%E6%BA%90%E5%91%BD%E5%90%8D%E7%A6%81%E5%BF%8C/ Android
DQUERY - D-query #sorting #tree English Vietnamese Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the
转载:http://blog.csdn.net/wj_j2ee/article/details/8560132 java保留两位小数问题: 方式一: 四舍五入 double f = 111231.5585; BigDecimal b = new BigDecimal(f); double f1 = b.setScale(2, BigDecimal.ROUND_HALF_UP).doubleValue(); 保留两位小数 -----------
洛谷P1440 求m区间内的最小值 ............................................................................... 以上代表我此时的心情,调了一个小时....只因为顺序,维护一个单调递增队列就好了,这里n很大,输出要优化,这才挽救了30分.. #include<bits/stdc++.h> using namespace std; int n,m; ],q[]; int top,tai; void Cin(int