【面试系列】-腾讯omg-测试开发-面试题---not finished！

1、mysql数据库复合索引是(a,b,c)，按照abc顺序建立复合索引idx_a_b_c,以下查询语句中使用到索引idx_a_b_c的有哪些？
A、(a) B、(b) C、(b,c) D、(a,b) E、(a,b,c)
解答：B、C
复合索引:Mysql从左到右的使用索引中的字段，一个查询可以只使用索引中的一部份，但只能是最左侧部分。例如索引是key index (a,b,c). 可以支持a | a,b| a,b,c 3种组合进行查找，但不支持 b,c进行查找 .当最左侧字段是常量引用时，索引就十分有效。

2、crontab定时任务：每周1到周5，下午1点和8点定时执行，怎么写？
解答：0 13,20 * * 1,2,3,4,5

3、下面C语言程序的打印结果是啥？
备注：涉及指针的一段c语言代码——–记不清了！

4、java代码片段，打印结果是啥？

for(i < a(); i < b(); i++){
    print("z");
}

public int a():
    print("x");
    return 0;
public int b():
    print("y");
    return 5;

解答：xyzyzyzyzyz
执行顺序是，先执行第一个表达式i < a();（这个只有在第一次循环的时候用到，后面的循环就不用了）。
接着判断第二个表达式i < b();，非零就执行循环体，零就退出，
接着运行循环体print("z");
最后运行第三个表达式i++（改变循环条件）

5、mysql的题目-查询表中所有工资大于本部门平均工资得员工的数量和部门名——-具体的表啥的记不清了
参考这里-几乎一样的题目。

【待解决】6、开发一个小程序，起3个线程，线程id分别为：A、B、C，每个线程打印10便自己的线程ID
小程序打印的最终结果是ABCABC……以此类推！
解答：请用python实现

7、读取文件f.txt，把重复的名字和重复数量，打印出来。
1,asdasd,39
2,asdasd1,39
3,asdasd,39
4,asdasd2,39
5,asdasd,39
6,asdasd3,39
7,asdasd2,39
8,asdasd1,39
解答：

#coding=utf-8
f = open("f.txt")
count_dict = {}
for line in f.readlines():
    name = line.strip().split(",")[1]
    if name in count_dict:
        count_dict[name] += 1
    else:
        count_dict.setdefault(name, 1)

for key,value in count_dict.items():
    print("%s:%d" % (key, value))

8、计算两个日期之间有多少天？
解答：此题不简单！注意考虑清楚再答题！
可以参考本人的github-这里

# By Websten from forums
#
# Given your birthday and the current date, calculate your age in days. 
# Account for leap days. 
#
# Assume that the birthday and current date are correct dates (and no 
# time travel). 
#
def judge_year(year):
    sumDaysYears = 365
    feb = 28
    if ((year % 4 == 0) and (year % 100 != 0)) or (year % 400 == 0) :
        sumDaysYears = 366
        feb = 29
    daysOfMonths = [ 31, feb, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
    return sumDaysYears, daysOfMonths


def sumDays(year, month, day):
    sumDaysYears, daysOfMonths = judge_year(year)
    if month >= 2:
        i = 0
        sum1 = 0
        while i < month-1 :
            #print i
            sum1 = sum1 + daysOfMonths[i]
            i += 1
        sumDaysFromOnetoMonthDay =  sum1 + day - 1
    else :
        sumDaysFromOnetoMonthDay = day - 1
    print sumDaysFromOnetoMonthDay, sumDaysYears - sumDaysFromOnetoMonthDay
    return sumDaysFromOnetoMonthDay, sumDaysYears - sumDaysFromOnetoMonthDay


def daysBetweenDates(year1, month1, day1, year2, month2, day2):
    ##
    # Your code here.
    ##
    sum1Year1, sum2Year1 = sumDays(year1, month1, day1)
    sum1Year2, sum2Year2 = sumDays(year2, month2, day2)
    bwteenYears = year2 - year1
    if bwteenYears < 0 :
        print "year2 must > year1"
    elif bwteenYears == 0 :
        return sum1Year2
    else :
        if bwteenYears >= 2 :
            i = 0
            sumTmp = 0
            while i < bwteenYears-1 :
                sumDaysWholeYear, daysOfMonths = judge_year(year1 + i + 1) 
                sumTmp = sumTmp + sumDaysWholeYear
                i += 1
            return sum2Year1 + sum1Year2 + sumTmp
        return sum2Year1 + sum1Year2
# Test routine

def test():
    test_cases = [((2012,1,1,2012,2,28), 58),
                  ((2012,1,1,2012,3,1), 60),
                  ((2011,6,30,2012,6,30), 366),
                  ((2011,1,1,2012,8,8), 585 ),
                  ((1900,1,1,1999,12,31), 36523)]
    for (args, answer) in test_cases:
        result = daysBetweenDates(*args)
        print result
        if result != answer:
            print "Test with data:", args, "failed"
        else:
            print "Test case passed!"

test()

【面试系列】-腾讯omg-测试开发-面试题---not finished！

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