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请编写函数fun(),它的功能是:求出1到1000之内能...

void fun(int *a,int *n) { int i, j, k; *n = 0; for (i=1,j=1,k=1; i

#define N 1000 #include"stdio.h" main() { int sum; sum=fun(7); printf("%d以内所有%d的倍数之和为:%d\n",N,7,sum); getch(); yzj(); } int fun(int m) {int i,sum=0; for(i=1;i

void fun(int *a,int n,int *k) { int i,j=0; *k=0; for(i=1;i

int fun(int a,int b) {int m,i=0,n[80]; for(m=a;m<=b;m++) {if ((m/7==0&&m/11!=0)‖(m/7!=0&&m/11==0)) {n[i]=m; i++;} } printf(“%d”,i) }

public class Test {public static void main(String[] args){for (Integer integer : fun()) {System.out.println(integer);}}public static List fun(){List list=new ArrayList();for (int i = 1; i < 1001; i++) {if(i%7==0||i%11==0)list.a...

#include #include void func(double *a,double *b) { double sum,cha; sum=*a+*b; cha=*a-*b; cout

分析: 1. 很简单的方法就是遍历,只要能被7整除和11整除又不被两者同时整除的数保存下来 2. 显然这个数要么是7的倍数要么是11的倍数且不是77的倍数实现: // 实现2方法public Integer[] getNums(){ List nums = new ArrayList(); // 遍历7的倍...

function fun(byval w as integer) dim i,s as long for i=1 to len(w) s=s+val(mid(w,i,1))^2 next if s mod 5=0 then fun=1 else fun=0 end if end function

#include int fun(int a[100]){ int i, n = 0; for (i = 1; i

Function s(ByVal a As Integer,ByVal b As Integer) As Integer Dim i As Integer i=1 s=0 do while i

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