编程,求1~10000之间所满足各位数字的立方和等于它本身的数

如果是85个数字满足的话,那么应该是100-999,不是1-999#include <stdio.h>#include <stdlib.h>//求被3整除且

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

利用奇数平方和公式：1^2+..+(2n-1)^2=(1/3)n(4n^2-1)1^2+2^2+...+n^2=n(n+1)(2n+1)/61^2+2^2+...+(2n)^2=2n(2n+1)(4n

#includevoidmain(){\x09//c语言编程计算1到200之间的奇数和与偶数和\x09inta,sum1=0,sum2=0;\x09for(a=1;a

#include#include#includeintmain(){inta,b,c;srand((unsigned)time(0));a=rand()%100+1;b=rand()%100+1;c=

#includevoidmain(){ints,i,sum=0;scanf("%d",&s);for(i=1;i再问：真的是这样么哥们急啊我也不会再答：运行结果：3391521273339455157

只会做第1个:PrivateSubCommand2_Click()Dimi%Fori=100To500x=i\100y=i\10Mod10z=iMod10Ifx+y+z=5ThenPrintiNext

#include#includevoidmain(){inti=1;intpsquare;while(1){psquare=i*i;i++;if(psquare>=1&&psqua

intn=1,sum=0;while(n

1、s=0fori=2to100step2s=s+iendfor2、s=0fori=1to100s=s+iendfor3、s=0fori=1to100step2s=s+iendfor4、s=1fori

inti,sum;i=1,sum=0;while(i

1、clears=1fori=1to9s=s+i/(i+1)endfor2、clears=0fori=10to100ifmod(i,2)=0s=s+iendifendfo

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

var  i,n,s:longint; c:array[1..10000]oflongint;functionsum(p:longint):longint;var&nbs

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

main(){ints=0,a=2;while(1){s+=a*a;if(a>=100)break;a=a+2;}printf("%d",s);}//如果有问题就是S的值可能要LONG吧,如果是LON

什么语言?下面是VF的*编程求1~300之间能同时被6和8整除的数的和s=0FORi=1TO300IFi%6=0.and.i%8=0s=s+iENDIFENDFOR?"1~300之间能同时被6和8整除

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

PrivateSubCommand1_Click()Fori=iTo100IfiMod5=0OriMod7=0Thens=s+iEndIfNextiPrint"能被5或都7整除的数之和为"&sEndS

intlen=0,results[256];for(inti=0;i