编写程序:计算1~100中即能被3整除,又能被7整除的所有数之和.
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intmain(void){inti,ans;For(i=1;i再答:�������ټ�һ��#include再问:лл
////////////////////下面是whileintsum=0;int_i=0;while(_i
s=0k=1whilei
#includevoidmain(){intsum=0;for(inti=1;i
PublicSubqqqqq()DimiAsIntegerDimjAsIntegerDimsAsIntegerDimCAsBooleans=0Fori=2To100C=FalseForj=2Toi-1
#include#includeboolIsprime(intnum){if(num再问:亲、我复制粘贴进C++程序、显示有一个错误、麻烦你在修改下把、感谢了、给你追分~再答:没错啊!你把错误贴上来
dimiaslongdimsumaslongsum=0fori=1to100step2sum=sum+inextiprintsum
语句自己找给思路.第一种:暴力运算1.申请2个变量jsum,osum分别放奇数和、偶数和.赋初值0.2.做个循环从1到100,变量用i2.1判断i是否为奇数,是的话jsum=jsum+i,否则osum
#include#includeintmain(){intsum=0;intindex=0;for(index=1;index
PrivateSubCommand1_Click()DimsAsString,nAsInteger,iAsIntegerFori=1To100n=n+iNextiPrint"1+2+3+……+100=
declare@iint,@Jsumint,@Osumintset@i=1set@Jsum=0set@Osum=0while(@i
intsum=0;//for循环for(inti=1;i
inti,S=0;for(i=1;i
dimxasinteger,dimsumasintegerforx=1to100ifxmod3=0thensum=sum+xendifnextxprintsum
publicclassTest{publicstaticvoidmain(String[]args){ints=0;intn=1;for(inti=0;i
用什么语言啊?如果是java的话就如下:ints=0;for(inti=1;i
VB的我知道PrivateSubForm_Click()DimsumAsIntegerFori=1To100IfiMod3=0Thensum=sum+iEndIfNextPrintsumEndSub单
;MOVAX,AANDAX,B;AX=aANDbMOVBX,AXORBX,B;BX=aXORbADDAX,BXADDAX,BX;AX=2*(aXORb)+aANDbADDAX,A;AX=a+2*(aX
CLEARSETTALKOffs1=0forn=1to10s1=s1+jc(n)next"1!+2!+3!+.+10!=",s1FUNCTIONjcPARAMETERSis=1FORj=1TOis=s
#includeintmain(){inti,s=0;for(i=1;i