c语言编写用通项公式前n项的和
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![c语言编写用通项公式前n项的和](/uploads/image/f/531575-71-5.jpg?t=c%E8%AF%AD%E8%A8%80%E7%BC%96%E5%86%99%E7%94%A8%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8C)
是前n个元素么?main(){inti=n,sum=0;max=a[0];min=a[0];for(i=0;ia[i])?max:a[i];min=(min再问:输出的avg不是整数再答:哎呀,忘了这
main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2
什么时候要?再问:呵呵,什么时候这个小问题解决了就不要了啊!·~~~~~~~~~~~~~~啊~开玩笑的啊,呵呵,对这个程序,当n一定时(比如20,30,40等等)我都可以写出来,但是.......当n
/*1-2/3+3/5-4/7+5/9-6/11+.的前n项之和*/intmain(intargc,char*argv[]){intnum_n=1;inti=0;doubleresult=0.0;pr
#includeintmain(){intn,i,t;floats,a;scanf("%d",&n);a=0;s=0;for(t=n;t>=1;t--){a=0;for(i=1;
#includeintmain(){inti,sum=0;for(i=1;i
main(){inta,sum,i;for(i>0;i
#include#includedoublefun(intn){doublesum=0.0;inti;intflag=-1;for(i=1;i{flag=(-1)*flag;sum+=1.0/i;}r
#include#definemaxsize50main(){intm,n,d,i,count;intA[maxsize];\x09printf("\n请输入n,m的值,以逗号分开:");\x09sc
//m与n的值不能太大,且用空格分隔.如:53#includeintfun(intn){\x09if(n==0||n==1)return1;\x09elsereturnn*fun(n-1);}void
#include <stdio.h>int calculate_mode(int number [],int n)//求众数{\x09in
#includemain(){inti,j,n,k,t;doublesum=2;printf("inputanumber:\n");scanf("%d",&n);j=2;k=1;for(i=1;i再问
sum=sum+1/(5*i+1);这一句,1/(5*i+1)的值是整数的,所以它一直是0这样好像可以sum=sum+(double)1/(5*i+1);
#includeintmain(intargc,char*argv[]){intm,n;inti;intsum;printf("Pleaseinputmandn:\n");scanf("%d%d",&
因为3个加法的运算规律是相同的,使用一个函数来计算循环值#includeintfun(intn)//计算累加结果函数{//这里还可以判断下n是否小于等于0intsum=0;inti;for(i=1;i
#includeintcal(intm,intn){intret=0;ret=m%n;returnret;}intmain(intargc,char**argv){intm,n,max,min
#include <stdio.h>int abc(int x,int y);void main(){int n1,n2,i;
#include <stdio.h>long u, v;void addrat(int, int);void lowterm()
#includedoublefun1(intn){inti;doublesum=0;for(i=1;i
#include"stdio.h"intmain(){\x05inti,j,fenzi;\x05floatresult=0,fenmu;\x05for(i=1;i01:-1;\x05\x05for(f