编写程序求出1-1000中能够同时被2与3整除的数并求它们的和 vb
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#includevoidmain(){longi;intj;printf("Pleaseinputnumber:");scanf("%ld",&i);for(j=999;j>=100;j--)if(i
include<stdio.h>int main(void){ int i= 0 , sum
供参考……#include"stdio.h"voidmain(void){intn=555555,i;for(i=999;i>99;i-=2)if(!(n%i)){printf("%d的
#includevoidmain(){intsum=0,i,temp;for(i=0;i
XOR\x09AL,ALMOV\x09CX,16Count:ROR\x09BX,1ADC\x09AL,0LOOP\x09Count或MOV\x09CX,BXJCXZ\x09BEYONDXOR\x09A
intsum=0;for(inti=10,i
#include <iostream>#include <string>#define M 4#define N 4
#include#includeintis_prime(intn);intmain(){inti,sum=0;for(i=2;i
inttemp1;intlength=0;intlength2=0;for(inti=0;i{if(i==0)temp=b[i];else{if(temp==b[i])length++;else{if
#includevoidmain(){ints=0;for(inti=0;i
#includeintmain(intargc,charconst*argv[]){intnumbers[20];inti,sum=0,count=0;for(i=0;i{scanf("%d",&nu
#include <stdio.h>int main() {int i = 1000;doif (555555
intcount=0;for(inti=1;i
#includeintfun();voidmain(void){inti,n,a[500];n=fun(a);for(i=0;iprintf("%d",a[i]);}intfun(inta[]){in
所有数位和:比如898=8+9+8=25,能被5整除自身能被三整除,比如111,111/3=37所有数位和能被5整除并且自身能被3整除的,比如555(ps:其实自身能被3整除的数,所有数位和也能被3整
#include#include#include"math.h"voidmain(){longinti,j=0,x;for(i=1;i
%主函数functiontest()x=[];fori=1:200flag=isprime(i);ifflagx=[x,i];endenddisp('1-200之间的质数有:');xend%子函数fu
给,已经编译运行确认:#include#includeintmain(){inti,sum=0;for(i=1;i500){sum-=i;break;}}printf("sum=%d",sum);ge
inputtoNFORi=1tonifmod(i,7)=0?iENDifENDFo
#includeintsum(intn){intj,s=0;for(j=1;j