private
sub
command1_click()
dim
sum
as
long
sum
=
0
for
i
=
1
to
1000
if
i
mod
5
=
0
and
i
mod
7
=
0
then
list1.additem
i:sum
=
sum
+1
next
msgbox
"1-1000中能被5和7同时整除的数有"
&
sum
&
"个"
end
sub
n=1
到
100
判断
n/3=整数
{
判断
n/5=整数
{i=i+1}
提示=提示&N&","
}
遍历结束
输出
总共有I个数
分别是<提示>
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还有运行图，希望对你有帮助。#include<stdio.h> main(){int i,n=0;printf("所有符合条件的数字如下:\\n");for(i=1;i<=500;i++)if((i%3==0)&(i%5==0)&(i%7==0)){n++;printf("%d ",i);} printf("\\n总共%d个\\n",n);} ...
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vb 程序:计算1-100之间所有能被3整除的数之和
include <stdio.h>int main(){int i, sum_3 = 0, sum_5 = 0;for (i = 1; i <= 100; ++i){if (i % 3 == 0){sum_3 += i;}if (i % 5 == 0){sum_5 += i;}}printf("1-100之间能被3整除的所有数字之和:%d\\n", sum_3);printf("1-100之间能被5整除的所有数字...
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const N = 100 dim i,iCount iCount =0 '计数 dim sum : sum =0 for i=1 to N if i mod 5 =0 or i mod 7 =0 then iCount = iCount + 1 textBox1.Text = textBox1.Text & cstr(i)sum = sum + i if iCount mod 4=0 then textBox.Text = textBox1.Text & vbcrlf ...
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Dim i As Integer, x As Long For i = 100 To 999 If i Mod 3 = 0 Or i Mod 5 = 0 Then x = x + i End If Next i Label1.Caption = x 请留意“能被3或者5整除”中的“或者”两字，应该用Or。如果是“能被3和5整除”才用And。另外x设为Integer有可能会溢出，应设为Long ...
