求几个数

Question

 上次考试出的题目，纠结了半天弄不出来，突然发现自己好笨的。 求出1~1000之间的所有能被7整除的树，并计算和输出每5个的和。

Answer 1

declare @index1 int
declare @tempindex int
declare @value1 int
set @index1=1
set @value1=7
set @tempindex=0
while(@index1<1001)
begin
if @index1%@value1=0
begin
print @index1
set @tempindex=@tempindex+1
print 'tempindex'+convert(nvarchar(36),@tempindex)
if(@tempindex=5)
begin
set @tempindex=0
print 'value'+convert(nvarchar(36),@index1+(@index1-7)+(@index1-14)+(@index1-21)+(@index1-28))
print 'value'+convert(nvarchar(36),@index1*5-@value1*10) 
end

end
set @index1=@index1+1
end/*
7
tempindex1
14
tempindex2
21
tempindex3
28
tempindex4
35
tempindex5
value105
value105
42
tempindex1
49
tempindex2
56
tempindex3
63
tempindex4
70
tempindex5
value280
value280
*/
if not object_id('tempdb..#testTb') is null
begin
drop table #testTb
end
create table #testTb
(
number int
)declare @index int
declare @value int
set @index=1
set @value=7
while(@index<1001)
begin
insert into #testTb
select @index
set @index=@index+1
endselect * from  #testTb where number%@value=0

Answer 2

int nCount = 0;
int nSum = 0;
for(int i=1; i<=1000; i++)
{
    if (i % 7 == 0)
    {
        nCount++;
        nSum += i;
        if (nCount >= 5)
        {
            Console.WriteLine(nSum);
            nCount = 0;
            nSum = 0;
        }
    }
}

Answer 3

List<int> list=new List<int>();
for(int i=7;i<=1000;i++)
{
    if(i%7==0)
        list.Add(i);
    if(list.Count%5==0)
        Console.WriteLine(list[list.Count-1]+list[list.Count-2]+list[list.Count-3]+list[list.Count-4]+list[list.Count-5]);
}

Answer 4

int a = 0, b = 1, c = 0;            for (int i = 1; i <= 1000; i++)
            {
                a = i % 7;
                if (a == 0)
                {
                    Console.Write(i + "+");
                }
            }
            Console.WriteLine();
            Console.WriteLine("---------------------华丽的分割线------------------");            for (int i = 1; i <= 1000; i++)
            {
                if (b != 6)
                {
                    c += i;
                    b++;
                }
                else
                {
                    Console.WriteLine(c);
                    b = 1;
                    c = 0;
                    c += i;
                    b++;
                }
            }

Answer 5

int nCount = 0;
int nSum = 0;
for(int i=1; i<=1000; i++)
{
    if (i % 7 == 0)
    {
        nCount++;
        nSum += i;
        if (nCount >= 5)
        {
            Console.WriteLine(nSum);
            nCount = 0;
            nSum = 0;
        }
    }
}赞同

Answer 6

 List<int> d7 = new List<int>();
            List<int> wsum = new List<int>();
            int nCount = 1, nsum = 0;            for (int i = 1; i <= 1000; i++)
            {
                if (i % 7 == 0)
                {
                    d7.Add(i);
                }
                if (nCount >= 6)
                {
                    wsum.Add(nsum);
                    nsum = i;
                    nCount = 2;
                }
                else
                {
                    nsum += i;
                    nCount++;
                }
            }
            Console.WriteLine("被七整除");
            foreach (int i in d7)
            {
                Console.WriteLine(i.ToString());
            }
            Console.WriteLine("每5个数相加");
            foreach (int i in wsum)
            {
                Console.WriteLine(i.ToString());
            }
            Console.ReadKey();

Answer 7

这样更容易理解。int nCount = 0;
            int nSum = 0;
            for (int i = 1; i <= 1000; i++)
            {
                if (i % 7 == 0)
                {
                    Console.Write(" {0}", i);
                    nCount++; nSum += i;
                    if (nCount % 5 == 0)
                    {
                        Console.WriteLine("={0}", nSum);
                        nCount = 0;
                        nSum = 0;
                    }
                }
            }
            Console.ReadKey();