编程求和S=1 2 3 - 100.用do while-loop实现
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![编程求和S=1 2 3 - 100.用do while-loop实现](/uploads/image/f/6764721-33-1.jpg?t=%E7%BC%96%E7%A8%8B%E6%B1%82%E5%92%8CS%3D1+2+3+-+100.%E7%94%A8do+while-loop%E5%AE%9E%E7%8E%B0)
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
1.阶乘函数:FunctionFactorial(nAsLong)AsLongIfn=1ThenFactorial=1ElseFactorial=n*Factorial(n-1)EndFunction
//VC6.0编译运行通过//求和s=1!+3!+5!#include//求阶乘函数intf(intn){if(1==n){return1;}returnf(n-1)*n;}//main函数intma
#includeintmain(){intt=1;inti;ints=0;for(i=2;i
有内部函数为factorial(n)或者n=20;y=1;fori=1:1:n;y=y*i;end;y以上是求阶乘求和为funticonfun(x,n)sum=0tem=1fori=0:ntem=i*
INPUT"输入自然数:"TOnnSUM(n)RETURNFUNCTIONnSUMPARAMETERSnns=0FORi=1TOnns=s+iENDFORRETURNs
的确错了-------------------s+=i*10+(++i)其实是s=s+【i*10+(++i)】,而(++i)则是i+1于是代码变成了for(inti=1;i
这个就是用for实现的#include <stdio.h>int main(){ int num = 0; 
因为Sn=12+34+58+716+…+2n−12n,所以12Sn=14+38+516+…+2n−32n+2n−12n+1,两式相减得:12Sn=12+24+28+216+…+22n−2n−12n+1
#includeintmain(){inti,sum=0;for(i=1;i
sum=1/5;fori=5:99sum=1/(i+1)+sumend
(1)题程序如下:settalkoffs=0i=1dowhilei
#includeusingnamespacestd;intf(intn){ints=0,t=1;for(inti=1;i
Dimi,sAsIntegerFori=2To100step2s=s+iNexti
#includeusingnamespacestd;intmain(intargc,char*argv[]){ints=0;inttemp=1;inti;for(i=1;i
var n,k,i:longint; x,p:extended;begin readln(k); x:=0;&n
把你的Pij矩阵告诉我,我来试试
用for循环计算inta=2;ints=0;for(inti=1;i
#include#includeintmain(void){\x05intm;\x05scanf("%d",&m);\x05while(m--){\x05\x05intn,i;\x05\x05doub
vara,b,c,x,y:integer;beginread(a)whilea0dobeginifa