编程计算s=1! 2! 3! ...... 10!(n!表示计算n阶乘)
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![编程计算s=1! 2! 3! ...... 10!(n!表示计算n阶乘)](/uploads/image/f/6764860-28-0.jpg?t=%E7%BC%96%E7%A8%8B%E8%AE%A1%E7%AE%97s%3D1%21+2%21+3%21+......+10%21%28n%21%E8%A1%A8%E7%A4%BA%E8%AE%A1%E7%AE%97n%E9%98%B6%E4%B9%98%29)
数学计算可以简化过程:因为1-2=-1,3-4=-1...所以得出-1+(-1)+...,重复(100/2)次,即代码为(这是最简单而且最快速的办法):DimsAsInteger=-(100/2)如果
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
clears=0m=0fori=1to22m=m+is=s+mendfor?"S=1+(1+2)+(1+2+3)+…+(1+2+3+4+…+22)=",s
s=0sum=1;for(i=1;i
programjuio;vari:integer;a,s:real;begina:=0;fori:=1to100dobegins:=a+i;a:=s;end;writeln(s:0:0);readln
PrivateSubForm_Activate()a5EndSubSuba(nAsLong)DimiAsLongFori=1TonSum=Sum+1/f(i)NextiPrintSumEndSubFu
s=0i=1dowhile.T.P=(1/(i*(i+1)*)s=s+1/pif1/p
i=0s=0DOWHILEi
PublicFunctions(n)Dimi,j,t,resFori=1Tont=1Forj=1Toit=t*jNextres=res+tNexts=resEndFunction
你好!PrivateSubForm_Click()DimiAsIntegerDimsAsIntegerFori=1To100s=s+iNextiPrintsEndSu
S=1!+2!+3!+…+n!,得n!=n*(n-1)*(n-2)*…*2*1
clearallcloseallx=1:1:20;lx=length(x);fork=1:1:lxs=0;s=s+factorial(x(k));end>>ss=2.4329e+018
#includeusingnamespacestd;intf(intn){ints=0,t=1;for(inti=1;i
1、clears=1fori=1to9s=s+i/(i+1)endfor2、clears=0fori=10to100ifmod(i,2)=0s=s+iendifendfo
PrivateSubCommand1_Click()DimiAsInteger,jAsInteger,nAsIntegerDimsum1AsLong,sum2AsLongn=inputbox("请输入
DimaAsIntegerDimsAsIntegerPrivateSubCommand1_Click()s=0Fori=1To100IfiMod2=1Thena=iIfiMod2=0Thena=-is
#includeusingnamespacestd;intmain(intargc,char*argv[]){ints=0;inttemp=1;inti;for(i=1;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
用for循环计算inta=2;ints=0;for(inti=1;i