求1-1000之间所有其个位数立方和等于该数的数(如:1.64.125.等等).
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intmain(void){inti;intsum=0;for(i=0;i
#include"stdio.h"voidmain(){inti;for(i=1;i
#includeintmain(){intn=0;for(inti=100;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
dimiasinteger,iiasintegerfori=1to1999if(imod3)=0and(imod7)=0andright(i,1)=3thenii=ii+1endifnextprint
#include"stdio.h"voidmain(){inti;for(i=1;i再问:(i%10==6)这个表示个位是6吗?再答:对的再问:如果要求把这些数都加起来应该怎么编?再答:#incl
PrivateSubCommand1_Click()Dimi,j,a,b,sFori=7To997a=FalseForj=1Toi-1IfiModj=0Andj>1Thena=TrueExitForE
#include#includeintmain(void){inti,j,n=0;intprime[1000];for(i=2;i
给你自然语言和PASCAL语言各来一段吧.PASCAL的vari,j,sum:longint;beginfori:=1to100doifimod2=1thenbeginwriteln(i);//不想换
#include <stdio.h>#include <math.h>#define N 1000void main(){&n
#include#includeintss(intn)/*检查n是否为素数,如果是则返回1,否则返回0*/{inti;for(i=2;i
ntmain(intargc,char*argv[]){charsound[10][5]={"yi","er","san","si","wu","liu","qi","ba","jiu","shi"}
1)令i=10;2)如果i>100则结束3)temp等于i乘以10加上64)如果temp对3求余等于0,则输出temp5)i等于i+17)转2)
settalkoffclearn=0forx=1000to9999a=int(x/1000)b=int((x-a*1000)/100)c=int((x-a*1000-b*100)/10)d=mod(x
5,6,25,76,376,625以下是c语言程序#includemain(){inti,k=10;longj;for(i=2;i
设个位数为X,十位数为X+2则7>X+2>55>X>3则X=4X+2=6这个两位数为62再问:能否再口语化一点?再答:因为“这两位数在五十和七十之间”,所以十位数(X+2)一定大于5小于7,ok?5>
设这个两位数为10a+b,a+3=b(a+10b)(10a+b)=574代入(11a+30)(11a+3)=574a=1所以b=4这个数是14
PrivateSubCommand1_Click()Sum=0Fori=0To9Print10*i+7;Sum=Sum+10*i+7NextiPrintPrint"Sum=";SumEndSu
#includemain(){longi,j,k;k=10;for(i=1;i再问:--------------------Configuration:Y17-Win32Debug----------
DimnAsInteger,sAsIntegerForn=100To999IfLeft(n,1)>Mid(n,2,1)AndMid(n,2,1)>Right(n,1)Thens=s+1NextPrin