(aaa....aa)n个a怎么编译?

运行正确,注意你输入的格式是：2,3.中间一定要有一个半角的逗号.再问：不对啊，输入2,3，输出的是245再答：再问：这是为什么再答：可能你的机器对浮点数的精度的处理比较特殊。。或者是我的机器比较特殊

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

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#includevoidmain(){inta,n,sum=0,temp=0;printf("inputtheaandn:\n");scanf("%d%d",&a,&n);printf

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

longsum;sum=0;inta,tmp;a=1;tmp=a;for(inti=0;i

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

PrivateSubForm_Click()DimaAsIntegerDimnAsIntegerDimsAsStrings=""Randomizea=Int(Rnd()*9)+1n=Int(Rnd()

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

stringgeta(intn){stringres="";for(inti=0;i

给你个最简单的：#include"stdio.h"main(){doublen1,x,t,t1;intcx,i;scanf("%lf,%d",&n1,&cx);t=n1;t1=n1;for(i=1;i

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

intfun(inta,intn){inti,r=0,t;t=a;for(i=1;i

/*2  (repeat=2)2 3 (a=2, n=3)8 5 (a=8, n=5)   

#includevoidmain(){inta,n,count=1;longintsn=0,tn=0;printf("pleaseinputaandn\n");scanf("%d%d",&a,&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

当i不等于n的时候在s字符串最后加上一个“+”号,当输出最后一个字符串,比如22222时候i=5,n=5,这个时候不要再s最后加上“+”号如果有这句,比如a=2,n=5时候输出是这样的：当a=2时,n

CLOSEALLCLEARALLCLEARcGetLenth_a=""nMaxLenth_n=09&&&n的最大数.***因VFP数字取值范围为：-2147483647到2147483646,即：**

#includevoidmain(){inta,i,k,sum,n,t=0;scanf("%d",&n);sum=0,i=1,k=2;do{t=t+k;sum=sum+t;k=k*10;i++;}wh

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.