变成寻找1000以内最小的10个素数与最大的10个素数，并计算20个素数之和

输出形式zuixiaosushu: zuidasushu: sushuzhihe:

10; i++) { printf("%d "stdio; j=sqrt(number); } printf("n";,a[i]); sum+=a[i];\nsushuzhihe:%d\.h> #include < } int main() { int i; int a[1000];\nzuidasushu:"); for(i=0; printf(",a[num-i-1]); sum+=a[num-i-1]; a[0]=2;math.h>zuixiaosushu:"); for(i=0; } } return flag; i< } printf(" } } int sum=0; int prime(int number) { int i,j,flag=1;10; i++) { printf("%d " for(i=2;i< i<=j;i++) { if(number%i==0) { flag=0; break; i++) { if (prime(i)) { a[num]=i; num++; int num=1; for (i=3; i<1000#include &lt,sum);

'添加一个textbox 单击窗体运行此程序 代码如下： private sub form_click() dim m as integer m = inputbox("请输入一个数的范围：") redim a(1 to m) as integer for i = m to 2 step -1 if prime(i) then text1.text = text1.text & i & " " b = b + 1 s = s + i if b mod 10 = 0 then exit for end if next print m & "个数中最大的十个素数之和是：" & s end sub function prime(a) as boolean for i = 2 to sqr(a) if a mod i = 0 then prime = false exit function end if next i prime = true end function