编程输入n计算公式sum

#include#includevoidmain(){intN,sum,i;sum=0;printf("N=");scanf("%d",&N);printf("sum=");for(i=1;i

privatesubform_click()dimnaslong,snaslongn=inputbox("请输入n")fori=0tonsn=sn+2*i+1nextmsgbox"1+3+5+...(

#includemain(){inta,n;doublesum=0.0;scanf("%d",n);for(a=1;a

#include"stdio.h"#defineN100voidmain(){inti,sum=0;for(i=2;i

你的程序中判断素数并加和的部分有问题在你的基础上修改如下PS:main不需要改动# include <stdio.h>int SumPrime(int&nbs

如果要求和的数据在A1：B20中,原来求和公式为=SUM(A1:B20)再改为=SUMPRODUCT(--A1:B20)试试.

main{inti,n,sum=0;cin>>n;if(i=0,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

定义unsignedintn,longlongintn1,计算过程用for循环每次*10,保存到n1,最后输出n1再答：算法思路大致就是这样了，具体代码应该不难写再问：再答：哦，原来是这个再答：那么把

PrivateSubCommand1_Click()DimnAsInteger,iAsIntegerDimsumAsDoublen=Val(InputBox("PleaseEnterN:"))sum=

importjava.awt.*;importjava.awt.event.*;classTestimplementsActionListener{TextFieldintext,outtext;Bu

#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i

#include#includeintmain(void){intn,m,i,j,t;scanf("%d%d",&n,&m);i=m>n?m:n;j=m>n?n:m;while(j){t=i%j;i=

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

#includeintmain(){inti;doublen,s=1,x=1,y=1;scanf("%lf",&n);for(i=1;;i++){x*=n;y*=i;s+=x/y;if(x/y

1-2=-1,2个数字产生1个-1,即,和为-11-2=-1,3-4=-1,4个数字产生2个-1,即,和为-21-2=-1,3-4=-1,5-6=-1;6个数字产生3个-1,即,和为-3n个数字产生n

sum代表总和,及∑

这不是数学么,总结出通项就行了.intn=5；//定义要循环的次数doublesum=0;for(inti=1;i

#include<stdio.h>int fx(int x,int y){int sum=0,a=x;;for(int i=0;i&l

publicclasssum{/***@paramargs*/publicstaticvoidmain(String[]args){//TODOAuto-generatedmethodstubintn