编程从键盘输入n,然后计算并输出n! = 1 × 2 × 3 ×

packagetest1;importjava.util.Scanner;publicclassTest6{publicstaticvoidmain(String[]args){Scannerinpu

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

intn,sum=0;cin>>n;for(inti=1;i

varch:char;//第一行输入的字符r,x,y,z,a,b:longint;//r为半径,x,y,z为三角形边长,a,b为长宽p:double;//p为海伦公式里的一个量beginreadln(

#includeusingnamespacestd;intmain(){inti,n,s=0;coutn;s=0;for(i=1;i

主函数这样intmain(){intsum=0;inti;intn;printf("Pleaseinputanumber:");scanf("%d",&n);for(i=1;i

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

如图：

#include#includeintfactorial(intn);doublefact(intn);doublepower(intn);intmain(){inti,j;\x05doubleff=

你好像对C语言的赋值符号不理解?e=(1000*a)+(100*b)+(10*c)+(1*d);这样是把(1000*a)+(100*b)+(10*c)+(1*d)的值赋给e,而不是把e分解成a,b,c

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

#includevoidmain(){longi,n,num=1,result=0;printf("Pleaseinputanum:");scanf("%ld",&n);for(i=1;i

#includevoidmain(){inta[7][4];intsum=0;inti,j;printf("请输入7*4矩阵：\n");for(i=0;i

#includevoidmain(){\x09inta[20],sum=0,count=0;\x09printf("请输入20个整数\n");\x09for(inti=0;i0)\x09\x09{\x

//给出三角形的三个边长a,b,c求三角形的面积//用海伦公式来计算三角形的面积p=(a+b+c)/2,S=根号下{p(p-a)(p-b)(p-c)}#include#includeintmain(v

#includemain(){inti,j,n,k,t;doublesum=2;printf("inputanumber:\n");scanf("%d",&n);j=2;k=1;for(i=1;i再问

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

PrivateSubCommand1_Click()DimiAsInteger,jAsInteger,nAsIntegerDimsum1AsLong,sum2AsLongn=inputbox("请输入

intmain(){intn=0;inti=0;intsum=0;scanf("%d",&n);for(i=2;i

1.input"请输入一个小于6,大于2的正整数："tonifn=6"输入数据不符合要求,请重新运行程序再输入!"elses=0k=0dowhilk

编程从键盘输入n,然后计算并输出n! = 1 × 2 × 3 × - × n.