C语言计算1 n*2 (n-1)*3 (n-2)*..

#includevoidmain(){intn;ints=0;printf("请输入一个正整数：");scanf("%d",&n);printf("该整数除1和其本身的因子为：\n");for(int

#include#includeintmain(){doublex1,x2,a;printf("pleaseinputanumble:");scanf("%lf",&a);if(a=1e-5);pri

#include#includevoidmain(){inti,j,n;intf,sum;f=1;i=1;sum=0;printf("请输入表达式n的值：\n");scanf("%d",

intfun(intn){//在该函数体中求1～n的累加值//用return语句返回累加值intsum=0;inti;for(i=1;i

#include"stdio.h"main(){intsum=0,j,i,k,n,p=1;scanf("%d",&n);for(i=1;i

你自己写点,有问题我帮你调试.这个不难再问：不会写啊再答：编程就是个熟练活儿，多练习就会了# include <stdio.h>int main(void){

#includeintn,i;doublesum;voidmain(){sum=0;scanf("%d",&n);for(i=1;i

这个东西关键是递归算法的确定,需要一点数学知识.#include//递归计算函数intcalc(intx){if(xif(x==1){return1;}elseif(x==2){return3;}el

ak=kC(k,n)=k*n!/k!*(n-k)!=n*(n-1)!/(k-1)!(n-k)!=nC(k-1,n-1)故原式=nC(0,n-1)+nC(1,n-1)+nC(2,n-1)+……+nC(n

longadd(intn){intt=n-1;if(t>1){longresult=n*t;longsum=result+add(t);returnsum;}else{returnn;}}楼上的方法,

三种写法都写了,自己慢慢体会吧,你要自己学会思考,尽量不要依赖答案.

#includeintmain(void){intn=0,i=0;doublep=1,q=0,s=0;printf("n=");scanf("%d",&n);for(i=1;i

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

#include<stdio.h>intmain(){longN,t=1,sum=0,i;label:printf("pleaseinputN(n>1&&n<1

#includevoidmain(){longi,n,num=1,result=0;printf("Pleaseinputanum:");scanf("%ld",&n);for(i=1;i

#includeintmain(){\x09intn;\x09unsignedinttmp=0;\x09printf("pleaseinputadata:");\x09scanf("%d",&n);\

#include"stdio.h"main(){inti,p=1,sum=0;for(i=1;i<=10;i++){p=p*i;sum=sum+p;}printf("%d\n",sum);}运行

每次从到for(t=1,s=0,i=1;t

#include <stdio.h>#include <string.h>main(){\x05int n=0;\x05int s=0;

#includeintmain(){intn,i;doublesum=0;scanf("%d",&n);for(i=1;i