c语言计算数列前n项和

#includemain(){inta=3,b=2,tmp_a,tmp_b,i;doublesum=0;for(i=0;isum+=1.0*a/b;tmp_a=a-b;tmp_b=a+b;a=tmp_

//求数列的前n项之和,数列初始元素为1/2//下一元素的分子是上一元素的分母//下一元素的分母是上一元素的分子分母绝对值之和//偶数项为负,奇数项为正doublesum(unsignedintn){

#include#includeintmain(void){intn;doublei,j,k;doubleres=0;printf("pleaseinputn:");scanf("%d",&n);i=

main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2

intsum(intm,intn){inti=m;intsum=m;if(m>n){printf("\nAmistake:m=%d>n=%d\n",m,n);return;}else{while(i{

#includeintmain(){intn,i,t;floats,a;scanf("%d",&n);a=0;s=0;for(t=n;t>=1;t--){a=0;for(i=1;

#include#defineN20main(){intf(intn);inti;for(i=1;i

doubley=k=s=t=1;这个分开来写：doubley=1,k=1,s=1,t=1;

#includeinta[100]={0};intfbnq(intn){intiRet=0;if(n>2){iRet=fbnq(n-1)+fbnq(n-2);}elseif(n==2){iRet=fb

#include#defineCOL10//一行输出10个longscan(){//输入求fibonacci函数的第N项intn;printf("InputtheN=");scanf("%d",&n)

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

解题思路：数列前n项和解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.

楼主这个百度有很多的,在此借用一下夜游神小翠的程序：#include#defineN20intFibonacci(intn){if(n==1||n==2)return1;elseretur

#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]

#include#include#defineN200longintF[N];voidFi(intn);longintSum(intn);voidmain(){intn;longintsum;prin

for(i=1;1再问：为何死循环了再答：1不是一直小于n....,应该是i小于n把再问：我去。。。打错字母了╮(╯▽╰)╭想打i的。。。

#includeintFib(intm){if(m==1||m==2)return1;returnFib(m-1)+Fib(m-2);}voidmain(){intn,i;printf("请输入n的值

解题思路：前n项和,错位相减解决问题解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include

经鉴定,LZ对于C语言中的数组/指针以及函数的相关知识点还没掌握好这一题的正确思路应该是,设计一个函数intfeibo(intn),该函数的作用是返回斐波那契数列第n项的值.主函数中获取用户输入的N,

#includevoidmain(){intn,i=1;doublea=0,x;printf("请输入要求X的前几项之和：");scanf("%d",&n);printf("请输入x的前m项(m>=n