读入一个正整数n,计算并输出前n项之和
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#include <stdio.h>void main(){int i,j,a,n,k=0,out=0;printf("请输入a与n:")
#include <stdio.h>int main(){\x09int n;\x09int nConut = 0;\x09sc
#includeintmain(){inti,k,n,t;doubleans;scanf("%d",&n);ans=0;t=1;k=1;for(i=1;i再问:输入输出示例Entern;3sum=0.
程序中的ab不知道做什么用的.pow(x,y)中x,y值反了,跟踪发现c=pow(x,-1)值为0这里因为c是整型,自动忽略小数位,值为0所以sum=sum+a*c;得到sum=0恒成立#includ
#includeintmain(){intn,num,max,min;scanf("%d",&n);for(inti=n;i>0;i--){scanf("%d",&num);if(i==n)//初始化
#includeusingnamespacestd;intmain(){inti;doublee,n,sum;cin>>n;for(i=0,e=-1,sum=0;i{e*=-1;sum+=e*1/(2
#includeintmain(){inta;intsum=0;scanf("%d",&a);if(a
#includevoidmain(){printf("pleaseinputanumber:");intnumber,temp=1;doublesum=0;scanf("%d",&number);fo
#include#includevoidmain(){char*p[10]={"zero","one","two","three","four","five","six","seven","eight
条件肯定少了……intfun(inta[],intn){if(n==0)return……if(n==1)return……if(n==2)return……if(n>=3)returnfun(a,n-3)
#includevoidmain(){intn;inta[6][6];inti,j,sum=0;printf("inputn\n");scanf("%d",&n);printf("inputn*nma
OptionBase1PrivateSubCommand1_Click()n=InputBox("请输入一个正整数")Fori=1ToLen(n)Sum=Sum+Val(Mid(n,i,1))Next
楼上的看清楚了,人家要的是程序方法一:modicomma.prg然后在出现的程序框中输入input"n="tonx=0form=1tonx=x+mendfor"n=",x按CTRL+W保存,然后在命令
如果是要求前N个偶数之和:#includeusingnamespacestd;longFunctionadd(intn){longsum=0;for(inti=0;i
importjava.util.Scanner;publicclassOushu{Oushu(){try{System.out.println("请输入n值");Scannerin=newScanne
#includevoidmain(){intn,i=1,fa=1;doublesum=1;scanf("%d",&n);for(i=1;i
我简单看了看哈,嘿嘿您个瞧瞧printf("Numberofdigit2:%d\n",number);这句,number都还没有赋值呢~~函数里面的变量number在函数完成时就死掉了...和你这个n
PrivateSubCommand1_Click()Sum=1Fori=1ToText1.TextSum=Sum*iNextiPrintSumEndSub再问:那在窗体上输出九九乘法表??你会吗??真
直接读int不好吗inta,b;a=reader.nextInt();b=reader.nextInt();再问:你看好题,数字是由英文给出的再答:那按“”分割,split("")