编程计算1! 2! .... n!-搜答案-智慧作业帮

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

N>200也就是说最后一个是201*202,总共是201项相加.movbx,1movdx,0movah,2movcx,200s:moval,ahincahmovsi,axmulahaddbx,axad

你自己写点,有问题我帮你调试.这个不难再问：不会写啊再答：编程就是个熟练活儿，多练习就会了# include <stdio.h>int main(void){

longfunc(intm){inti,sum=1;for(i=1;i

#includevoidfun(long*num,intnn);intmain(){intn;longsum=0;cout再问：调试了好像有点问题再答：#includevoidfun(long*num

楼主这个百度有很多的,在此借用一下夜游神小翠的程序：#include#defineN20intFibonacci(intn){if(n==1||n==2)return1;elseretur

#include<stdio.h>intmain(){longN,t=1,sum=0,i;label:printf("pleaseinputN(n>1&&n<1

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

#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]

#includelongcountNum(intn){longnum=1;intm=n;while(m){num*=m--;}returnnum;}longcountSum(intn){longi=0

functionCalculateE(n:integer):real;//计算n项,返回e的近似值varm,e:real;//为了防止阶乘太大而溢出,把m改用real变量.i:integer;begi

publicf(byvalnasinteger)aslongifn=1thenf=1elsef=f(n-1)+nendifendfunction

#includemain(){doublesum=1;//这么做是为了减少一次循环intn;for(n=1;n

S=1!+2!+3!+…+n!,得n!=n*(n-1)*(n-2)*…*2*1

#includeusingnamespacestd;intf(intn){ints=0,t=1;for(inti=1;i

privatesubcommand1-click()dimn%,y#,s&n=2:y=0:s=1dos=s*ny=y+1/sn=n+1loopuntil1/s

不需要用pow的double expx(double x){ double ret = 1;

clearinput"n="tons=1fori=1tons=s*iendfor"n!=",s

用for循环计算inta=2;ints=0;for(inti=1;i

PrivateSubForm_Click()DimsAsLongDimt1AsLongs=0Fori=1Tont1=1Forj=1Toit1=t1*jNexts=s+t1NextPrintsEndSu