从键盘输入n和字符ch,输出n行由ch构成的直角三角形.

importjava.util.Scanner;publicclassSum{publicstaticvoidmain(String[]args){Scanners=newScanner(System

#includevoidmain(){\x09intn,i,j,t,a[10000];\x09printf("请输入正整数N:");\x09scanf("%d",&n);\x09printf("请依次

#include#includeusingnamespacestd;intmain(intargc,char*argv[]){inti;couti;cout

#include#includeintmain(){doublex,f,s,t;intn;printf("pleaseinputx(0toexit):");scanf("%lf",&x);//doub

#includeintmain(){inta;intsum=0;scanf("%d",&a);if(a

VC6.0调试通过了,while循环结构#include"stdio.h"voidmain(){inti,sum=0,n;i=1;printf("Inputn:");scanf("%d",&n);wh

楼主你好!根据你要求实现如下#include<stdio.h>int fun(int n){ if(n>0)return n*fun(n-1)

我只会PASCALvarn,i,max,min,j:longint;beginreadln(n);max:=-maxlongint;min:=maxlongint;fori:=1tondobeginr

CLSINPUTni=1:s=40FORk=1TOnPRINTTAB(i+s);FORj=1TOiPRINT"*";NEXTjPRINTi=i+2:s=s-3NEXTkFORi=1TOnPRINTTA

#includeusingnamespacestd;voidmain(){intn,i,j=0,k=0,a[100],b[100];//a[100]用来存放奇数,b[100]用来存放偶数cout

//Note:YourchoiceisC++IDE#includeusingnamespacestd;longintSfun(constintN){longintsum=1;if(N==0||N==1

从键盘输入一个自然数N(

floatfun(intn){floatsum=0;inti;for(i=1;i

题目与程序的思路有一点点的偏差而已.程序是：n+(n-1)+(n-2)+(n-3)+...+2+1

#includeintmain(){inti=0;intj=0;intn=0;chara='A';while(1){printf("初始化n:");scanf("%d",&n);if(nn){brea

#includevoidmain(){chara,b;intresult;printf("输入两个字符：\n");scanf("%c\n%c",&a,&b);result=a-'0'+b-'0';pr

#includevoidmain(){inti,j,n;printf("请输入n的值：\n");scanf_s("%d",&n);for(i=1;i

调出来了：int[]b=newint[n];这个语句应该是上面的这个,要申请n个空间,不是n-1,否则会数组下标越界.运行后,先输入一个整数n,表示下面还要输入整数的个数,然后在输入n个整数,求这n个

#include"stdio.h"voidmain(){charch;scanf("%c",&ch);if(ch>='0'&&ch='a'&&ch='A'&&ch

C++版程序1、intsum=0;for(inti=1;i>n;for(inti=1;i