编写程序 求1到n中能被3或7整除的数之和

程序如下：#include#includevoidmain(){inti,n,sum=0;scanf("%d",&n);for(i=1;i{sum+=pow(i,i);}printf("%d",sum

#includevoidmain(){floatsum=0.0;intn=0,sign=-1;scanf("%d",&n);for(floati=1;i

楼上的两个答案都不对.都忘了将n-1.publicclassTestNFactorial{publicstaticvoidmain(String[]args){inttarget=9876;run(t

iTot=0Fori=1To100IfiMod5=0OriMod7=0TheniTot=iTot+iNextiPrintiTot结果是:1680【由于字数限制,去除了Sub,及变量定义】

#includevoidmain(){inti,a=0,sum=0;for(i=1;;i++){a+=1;sum+=a;if(sum>1000){printf("n:%d\nsum:%d\n",i,s

vars,i,n:long;beginwrite('N=');readln(n);s:=0;fori:=1tondos:=s+i;writeln(s);end.

#includevoidmain(){inti,j=0;for(i=1;i

classMAXn{publicstaticvoidmain(Stringargs[]){intk=1,sum=0;while((sum+=k)

varpp=0,kk=0functionf1(x){pp=x*xreturn(pp)}functionf2(y){kk+=yreturn(kk)}for(vari=1;i

#!/bin/bashread-p"Enteranum:"numsum=0foriin$(seq$num)dosum=$((sum+i))doneecho"Sumis$sum"

'编写程序,计算满足条件1的3次方加2的3次方到n的3次方

C言语……#includemain(){inti,k,j,sum,num;sum=0;j=1;printf("请输入一个整数：");scanf("%d",num);for(i=1;i

ints=1;intn=1;while(s

declare@numintdeclare@maxnumintdeclare@tempnumintSET@num=1SET@tempnum=1While@tempnum再问：验证：select1*3*

Console.Write("请输入n=");intn=Convert.ToInt32(Console.ReadLine());intsum=0;for(inti=1;i{sum*=i;}Consol

inputtoNFORi=1tonifmod(i,7)=0?iENDifENDFo

#includeintmain(){intsum=0;inti=1;intj=1;for(;i

#includeintfun(intn){returnn==0?1:n*fun(n-1);}intmain(void){inti,sum=0;for(i=1;i

#includeusingnamespacestd;intmain(){intn,m,num=0,a,b,c;cin>>m>>n;while(m0){a=m%10;s=s+a;m=m/10;}m=c;

settalkoffclearinput"输入N值："tons=0FORi=1TOna=1FORj=1TOia=a*jENDFORs=s+aENDFOR"1!+2!+3!+...+n!=",ssett