编写程序完成一下数字金字塔

#includeintmain(){intx;inta[20];scanf("%d",&x);inti;for(i=0;x>=10;i++){a[i]=x%10;x=x/10;}

n=""@2,10say"计算n个数的平均值,先输入数值个数n"getnpicture"999999"readn=val(n)DIMENSIONx(n)store0toxs=0fori=1ton@4,

Stringresult=sentence.replaceAll("a","1").replaceAll("b","2").;依此写这个可以查API的

有个缺陷,如果是thisisatest,则把a也大写了,不知道这样是不是符合你的要求.（按你的题目讲是正确的）如果不符合,则稍加个判断即可.#include#includeintmain(){char

#include <stdio.h>int moreThanAvg(float score[]){    int&nb

varn,i,min,max,maxi,mini,s:integer;x:array[1..100]ofinteger;ans:real;beginreadln(n);fori:=1tondoread

代码如下,供参考：      program test      im

//#include"stdafx.h"//vc++6.0加上这一行.#include"stdio.h"voidmain(void){inti,j;for(i=0;i

intmain(intargc,char*argv[]){inti,b,s,g,sum=0;for(i=100;i

vari,j,n,m:longint;procedureprint(c:char;a:integer);vari:longint;beginfori:=1toadowrite(c);end;begin

publicstaticvoidmain(String[]args){for(inti=0;ifor(intt=i;t>0;t--){//输出前面的空格System.out.print("");}fo

System.out.println(1+""+2+""+3+""+4);System.out.print(1+"");System.out.print(2+"");System.out.print(

额、temp是中间交换的临时变量,我以为还比较大小呢.publicclassc{publicstaticvoidmain(String[]args){inta=1;intb=2;inttemp;tem

miamia.

#include"stdio.h"#include"stdlib.h"intmain(){intx=3829;inta1,a2,a3,a4;intsum;a1=x/1000;a2=x/100%10;a

settalkoffcleainput"inputn(1-9):"tonFORi=1TOnFORj=1TOn-i''ENDFORFORj=1TOiSTR(j,1)ENDFORFORj=i-1TO1ST

PrivateSubCommand1_Click()Dimi,jAsIntegerFori=1To13Forj=1ToiPrint"";NextjForj=iTo13-i+1PrintChr(56-i

for(inti=0;ifor(intj=i;jSystem.out.print("");}for(intj=1;j0;j--){System.o

以下用的是C#编写金字塔程序doublerealCoord,imagCoord;doublerealTemp,imagTemp,realTemp2,arg;intiterations;for(imag

#include#include#includeintmain(){longi,a,n,len=0,sum=0;charstr[10];scanf("%ld",&n);sprintf(str,"%ld