编程求1~n中能被3或7

#include/*用for循环实现*/voidmain(){\x09intn;\x09intm;\x09printf("pleaseinputn\n");\x09scanf("%d",&n);\x0

n=val(inputbox"请输入一个数字.")fori=1tona=a+inextiprinta

//所谓模块化,说得通俗点,就是不断地嵌套调用函数来实现.且看下面：#includeusingnamespacestd;intf(intn)/*定义递归地求阶乘的函数f*/{if(n==0)retur

intmain(void){\x09intsum=0,i;\x09for(i=1;i

int sum(int n){    List<int> list = new List

.staticints;intn;s=1;..for(n=1;s

程序的实现的是这样的,先设置一个文本框,用于n值的出入.再设置一个命令按钮用于求解.代码如下：PrivateSubCommand1_Click()DimnAsInteger,iAsInteger,mA

n以内5或7的倍数之和:PrivateSubCommand1_Click()n=Val(Text1.Text)Fori=5TonIfiMod5=0OriMod7=0ThenSum=Sum+iNextM

#include"不同软件头文件不一样"main(){inti,n,s=0;scanf("%d",&n);for(i=1;i

#includeintmain(){\x09inti,n;\x09floatsum=0;\x09printf("请输入n：\n");\x09scanf("%d",&n);\x09for(i=1;i

for(x=n;x>=3;x--)'判断的最小值为3,小于3的值就没有意义{if(!(xmod3)or!(xmod7))'被3除无余数或被7除无余数count+=x;}x=n;while(x-->=4

for循环修改下fori=1to2n-1step2a=a+inexti

PrivateSubForm_Click()DimHeAsLongDimIAsInteger,MAsInteger,NAsIntegerM=Int(Val(InputBox("M的值","输入",3)

PrivateSubCommand1_Click()Dimi,sAsIntegers=0Fori=1To100IfiMod3=0OriMod7=0Thens=s+1EndIfNextiPrintsEn

programt1;vari,j,k2,k5,m,n:longint;beginreadln(n);fori:=1tondo{每个数的质因数2和5的个数的循环}beginm:=i;whilemmod2

#includeintmain(){inti,n,sum=0,flag=0;printf("请输入n:");scanf("%d",&n);for(i=1;i=100&&flag==0){printf(

inputnS=0Fori=1tons=s+i+iENDFORss=SQRT(S)

#includeintmain(){intn,i;longa,sum=1;printf("请输入一个奇数:\n");scanf("%ld",&n);for(i=3,a=1;i

#includemain(void){\x05inti,s;\x05\x05for(i=1,s=0;;i++)\x05{\x05\x05s+=i;\x05\x05if(s>=100)\x05\x05{

PrivateSubCommand1_Click()DimnAsInteger,iAsIntegerDimxAsVariant,sumAsDoublen=20x=CDec(x)x=1Fori=1Ton