编写求t=a...aaa的值,且a为1到9之间的数

l#include"stdio.h"intmain(){longi,j,a,n,sn=0,sum=0;/*sn为每一个a的值,sum为总值*/scanf("%d%d",&a,&n);for(i=1;i

#include#includemain(){inta,n,i,sum=0,sum1,sum2,sum3;//sum为最后结果printf("inputthenumbera:");scanf("%d"

DimaAsInteger,nAsInteger,iAsInteger,sAsDoublea=Val(InputBox("a=?"))n=Val(InputBox("n=?"))Ifa>0And

#includeusingnamespacestd;//Sn=a+aa+aaa+...+(n个a)//uA表示a//uN表示n//返回值为SnunsignedintSigmaN(unsignedint

#includemain(){intn;longa,sum=0;printf("pleaseinputaandn,andpressEntertocontinue\n");scanf("%ld%d",&

孩子,作业要自己做.可以写两层循环或者把aaaa写成一个函数,然后在加法程序里调用这个函数

我也不懂

vara,n,i,j:Integer;s:Int64;beginReadln(a,n);s:=0;if(a

inputa,ni=0m=0s=0DOb=10^im=a*b+ms=s+mi=i+1LOOPUNTILi>=nprintsend童鞋你不会是八中的吧.同一天同一道题.求班级,我就只给你一种解法吧再问：

设：x1=ax2=aaxn=aaaaa.a(n个)观察可得：x(n+1)=10xn+a等式两边同时加a/9得：x(n+1)+a/9=10xn+10a/9即：x(n+1)+a/9=10(xn+a/9)这

设共有n个数s=a+(10^1+1)*a+(10^2+10^1+1)*a+…+(10^n+10^(n-1)+…+10+1)*a=(10^n+2*10^(n-1)+…+(n-2)*10^2+(n-1)*

LZ不给分加一下最佳总可以吧?注意是BASIC不是BUSICinputa,ni=1s=0dos=s+aa=a+a*10^ii=i+1whilei

#include"stdio.h"main(){longa,i,p,s,N;printf("请输入a值:\n");scanf("%ld",&a);printf("请输入a小于等于:\n");scanf

PrivateSubCommand1_Click()Dima%,n%,i%,Sn$n=InputBox(请输入n的值)Randomizea=Int(9*Rnd+1)PrintTab(30);"n="&

#includeintNum(inta,intn)//Num函数用来求出aa...a(n个a){inti,sum=0;for(i=0;i

#include"stdio.h"#include"math.h"voidmain(){inta,n,i;longsum=0;printf("Pleaseinputa(0

#include#includevoidmain(){int*a,n,b;inti,j;printf("n?\n");scanf("%d",&n);printf("a?\n");scanf("%d",

是不是这个题目啊：编程计算a+aa+aaa+…+aa…a〔n个a〕的值,n个a的值由键盘输入#include#includevoidmain(){inta,n,sum=0,temp=0;printf(

c=0这个赋值要放到while（n>=1）里面,你可以放到i=n下面一行,你求c=c+b的时候每求完一次要重新初始化.

n[10^(i-1)]*a+s