编程,求1² 2² 3² .....n²的和vb
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clears=1input"请输入n的值:"tonfori=2tonk=(-1)^im=k/is=s+mendfor"s=1+1/2-1/3+1/4+...+1/n",s
n=val(inputbox"请输入一个数字.")fori=1tona=a+inextiprinta
#includeintmain(){inti,n=1,sum=0;for(i=1;i
publicclassTest{publicstaticvoidmain(String[]args){System.out.println(show());}publicstaticdoublesho
#include<stdio.h>void main(){ int sum=0,t=1,i; &nb
DimsAsSingleDimiAsIntegerFori=1To50s=s+i/(i+1)NextMsgBox"前50项和为"&s
#include<stdio.h>void main(){int i,sum=0;for(i=1;i<101;i++)sum+=i;printf("%d
#includevoidmain(){inti,sum;for(i=1,sum=0;i
inta=1,sum=0;voidmain(){for(i=1;i
程序的实现的是这样的,先设置一个文本框,用于n值的出入.再设置一个命令按钮用于求解.代码如下:PrivateSubCommand1_Click()DimnAsInteger,iAsInteger,mA
#include"不同软件头文件不一样"main(){inti,n,s=0;scanf("%d",&n);for(i=1;i
#includeintmain(){\x09inti,n;\x09floatsum=0;\x09printf("请输入n:\n");\x09scanf("%d",&n);\x09for(i=1;i
clears=0input"n="tonfori=1tonk=1forj=1toik=k*jendfors=s+kendfor?"s=1!+2!+3!+4!……+n!=",s
for循环修改下fori=1to2n-1step2a=a+inexti
programt1;vari,j,k2,k5,m,n:longint;beginreadln(n);fori:=1tondo{每个数的质因数2和5的个数的循环}beginm:=i;whilemmod2
1、clears=1fori=1to9s=s+i/(i+1)endfor2、clears=0fori=10to100ifmod(i,2)=0s=s+iendifendfo
inputnS=0Fori=1tons=s+i+iENDFORss=SQRT(S)
整个程序代码如下:#include"stdio.h"main(){inti,j;longa,total=0;for(i=1;i
PrivateSubCommand1_Click()DimnAsInteger,iAsIntegerDimxAsVariant,sumAsDoublen=20x=CDec(x)x=1Fori=1Ton
#includeintmain(){inti,j=1;doublesum=0.0;doublek;for(i=1;i