编程实现1到100之间所有的偶数的和-搜答案-智慧作业帮

你一点都不做?做一做大家给你修改才是正途

你好,可以这样编程]#includeintmain(){inti,sum=0;//定义循环变量和存放求和的变量sumfor(i=1;i

settalkoffclearfori=1to100ifi%3=0.and.i%7=0?iendifendforsettalkon

#includemain(){inti,sum=0;for(i=0;i

1su=0;fori=1:100ifmod(i,5)==0su=su+i;endendfprintf('所求和为：%d',su);2a=1;b=10;disp('x(1)=1\nx(2)=10');f

#includeusingnamespacestd;intmain(){intsum=0;for(inti=1;i

现在命令窗口输入：modifycommandqh让后在弹出了窗口中输入：clearstore0tosfori=8to100ifmod(i,8)=0s=s+iendifendfor"10到100之间的所

代码如下:运行条件:form1上有一个文本框,名称text1,一个按钮,名称command1运行结果显示在text1里面PrivateSubCommand1_Click()DimiAsIntegerM

#includeintmain(void){inti,sum;sum=0;for(i=1;i

PrivateSubCommand1_Click()Fori=3To100Forj=2ToSqr(i)IfiModj=0ThenExitForNextIfj>Int(Sqr(i))Thens=s&i&

#includevoidmain(){intsum=0,i;intoushu=0,jishu=0;for(i=0;i

#includeintmain(){\x09inti,sum=0;\x09for(i=1;i

voidmain(){intn=100;inta;intb;intc;for(n=100;n<=999;n++){a=n/100;b=n/10%10;c=n%10;if(n==(a*a*a+b*

clears=0fori=1to100ifi%30ifi%2=1s=s+iendifendifendfor"1至100之间非3的倍数的所有奇数之和为：",s

#include<stdio.h>main(){ inti=100; &

PrivateSubCommand1_Click()Fori=100To999IfIsPrime(i)Thenn=n+1Printi;IfnMod10=0ThenPrintEndIfNextiEndS

publicclassSumOfEven{\x09publicstaticvoidmain(String[]args){\x09\x09intsum=0;\x09\x09for(inti=1;i

#includeintmain(){inti;intsum=0;for(i=1;i{if(0!=i%7){sum+=i;}}printf("sum=%d\n",sum);return0;}

LOCALlnI,lcZf,lnGw,lnSw,lnBwFORlnI=100to999lcZf=ALLTRIM(STR(m.lnI))lnGw=VAL(SUBSTR(m.lcZf,3,1))lnSw=

fori=100to200ifint(i/3)=i/3andint(i/7)=i/7)thenprintinext