一到一百阶乘的和

#includeunsignedlongintjiecheng(intk)//阶乘函数{unsignedlongintn=1;while(k){n=n*k;k=k-1;}returnn;}unsign

100以内的质数共有25个,这些质数我们经常用到,可以用下面的两种办法记住它们.?一、规律记忆法?首先记住2和3,而2和3两个质数的乘积为6.100以内的质数,一般都在6的倍数前、后的位置上.如5、7

20的阶乘已经远远超过整数的表示范围.再问：那怎么办呢？再答：#includevoidmain(){doublei,n,sum;n=1;sum=0;for(i=1;i

Limn->无穷1!+2!+3!+n!/n!=1+1/n+1/[n(n-1)]+1/[n(n-1)(n-2)]...+1/n!=1

1one2two3three4four5five6six7seven8eight9nine10ten11eleven12twelve13thirteen14fourteen15fifteen16six

onetwothreefourfivesixseveneightnineteneleventwelvethirteenfourteenfifteensixteenseventeeneighteenni

#include"stdio.h"voidmain(){longa=1,n,s=0;for(n=1;n

初number['nʌmbə]数字one[wʌn]一two[tu:]二three[θri:]三four[fɔ:]四five[faiv]五six[siks]六se

1one2two3three4four5five6six7seven8eight9nine10ten11eleven12twelve13thirteen14fourteen15fifteen16six

质数又叫素数,从一到一百共有25个!2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97

sum赋初值0,还有一些语法错误修改一下,如下#includeusingnamespacestd;intmain(){inti,j,sum;sum=0;//加这句for(i=1;i

13位,long型应该能放得下.有的编译器可能放不下还有一个数据类型是longlong型解决的办法,要是放不下,只能用字符串来解决自己编写字符串的加乘运算.也不是很难再问：不会字符串的，跪求。。。再答

9.4269E+157

PrivateSubCommand1_Click()n=Val(InputBox("n="))t=1Fori=1Ton t=t*i s=s+tNextiPrin

一看你的帖子我有点晕的.一到十：One,two,three,four,five,six,seven,eight,nine,ten十一到二十：eleven,twelve,thirteen,fourtee

没有封闭的通项公式.

longfact=1,sum=0;应该改为longdoublefact=1L,sum=0L;或者fact和sum改为double型；longdouble为常双精度型；你的结果有19位,long是不行的

已通过测试,#includevoidmain(){doublesum=0,n=1;inti;for(i=1;i

1one2two3three4four5five6six7seven8eight9nine10ten11eleven12twelve13thirteen14fourteen15fifteen16six

一到十：One,two,three,four,five,six,seven,eight,nine,ten十一到二十：eleven,twelve,thirteen,fourteen,fifteen,si