编写一个程序s＝a+aa+aaa+aaa--

#include <stdio.h>void main(){int i,j,a,n,k=0,out=0;printf("请输入a与n：")

inta,n;doublesum=0;scanf("%d%d",&a,&n);for(sum=a,i=2;i

INPUT"a=";aINPUT"n=";ntn=0sn=0i=1WHILEi

要写的话太麻烦了.我就告诉你基本思路吧数字a没关系但是个数n的话,s=0个位有n个as=s+na十位有n-1个as=s+10(n-1)a百位有n-2个as=s+100(n-2)a千位有n-3个a一直做

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",&

解题思路：同学你好，本题首先由已知得出a,b的值，再把分式化简代入求解解题过程：

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：

voidmain(){inta,n,i;inttemp,s;scanf("%d%d",&a,&n);if(a9||n

#includeintfn(inta,intn);intmain(){inta,n,m,sum;printf("请分别输入a和n,用空格分开");scanf("%d%d",&a,&n);m=n+1,s

用你那个问题的格式：reada,nt←as←0forifrom1tons←s+tt←t*10+aendforprintsend

packageaddtest1;publicclassaddTest{publicstaticvoidmain(String[]args){inta=2;inttimes=6;intresult=0;

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

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C语言版：includeincludevoidmain(){inta,n,S;inti;printf("a=?\n")scanf("%d",&a);printf("\nn=?\n")scanf("%d

#include"stdio.h"longf1(longa,longn){if(n==1)returna;elsereturn(10*f1(a,n-1)+a);}longf2(intn){longr=

这是c++程序的源代码,我已经运行过了,准确无误#includeusingnamespacestd;intmain(){inta,n,u,sum;coutn;u=sum=a;cout