package suanfa; import java.util.Random; public class RandomStr { public static String getRandomStr(int length) { String base = "abcdefghijklmnopqrstuvwxyz0123456789"; int randomNum; char randomChar; Random random = new Random(); // StringBuffer
import randomcount=int(input('请输入你所想要手机号数量:'))prefix='138'for i in range(count): num=random.sample(range(9999999, 100000000), 1)#这样就避免了重复的号码 num1=prefix+str(num) print('%s'%num1)
Oracle 随机获取N条数据 当我们获取数据时,可能会有这样的需求,即每次从表中获取数据时,是随机获取一定的记录,而不是每次都获取一样的数据,这时我们可以采取Oracle内部一些函数,来达到这样的目的1) select * from (select * from tablename order by sys_guid()) where rownum < N; 2) select * from (select * from tablename order by dbms_random.va
//在两个数成对出现的数组中找到一个单独的数.比如{1,2,3.3,1,4.2},即找出4 #include <stdio.h> int find(int arr[], int len) { int i = 0; int ret = 0; for (i = 0; i < len; i++) { ret = ret^arr[i]; } return ret; } int main() { int arr1[] = { 1, 2, 2, 3, 1, 5, 3 }; int arr2[] =
这是一道面试题可以用以下的一些方案.第一种是很容易想到的采用循环的方式并且与1进行位与运算,具体代码如下. 1unsigned int GetBitNumOfOne_ByLoop1(unsigned int nValue) 2{ 3 const unsigned int nNumOfBitInByte = 8; 4 unsigned int nBitMask = 1; 5 unsigned int nBitNum = 0; 6 for(unsigned int i = 0 ; i < size