编写程序计算下列公式的值:y=x-想x 3!

#include#includeusingnamespacestd;intmain(){doublex,y;cin>>x;if(x

#includeusingnamespacestd;doublefun(doublex){\x05if(x>-10&&x-5&&x5&&xx;\x05\x05cout再问：麻烦用C语言写一下再答：#i

#include#includevoidmain(){doublex;doubley;coutx;if(-2

doublef(doublex){doubles=.0;if(x

#include "stdio.h"int fun(int n){   int sum=0,m=1,t=0;&

多绕几圈,可以实现的.>>result=maple('evalf','(Ei(1,y))')result=Ei(1,y)>>y=2y=2>>result=subs(result)result=Ei(1

#include<stdio.h>int main(){\x09int x,y;\x09printf("输入x的值："); \x09sca

#includedoublefun(intn){intr=1,i;if(n==0)return1;for(i=1;i1e-20){n=1/fun(i);e+=n;i++;}printf("e=%.6f

main(){longi,t=1;doublex,e=1.0;scanf("%f",&x);for(i=1;i

#includeusingnamespacestd;intmain(){intcount=1;intn;cin>>n;doublesum=0;for(inti=(2*n+1);i>0;i-=2){if

#include#includeintmain(){doubled;scanf("%lf",&d);printf("|y|=%.2lf\n",fabs(d));return0;}

ifx>=0theny=xelsey=-xendifprinty

#includemain(){intn,i;doublet,sum;/*1*/printf("请输入n的值\n");scanf("%d",&n);sum=2;i=1;t=2;/*2*/while(i

#includevoidmain(){floata,x,z;scanf("%f,%f",&a,&x);printf("pleaseinputa,x:,.\n");z=1.0/2*(a*x+(a+x)/

#include//希望对您有用#includeintmain(){intx,y;printf("Pleaseenterthenumber:");scanf("%d",&x);if(x

选Cs初始值为0,t初始值为1l从1递加到10你列举几个循环就知道了：从l=1开始,t:=1*1s:=0+1*1然后是l=2,t:=1*2,s:=1+2*1接着l=3,t:=2*3(2*3也就是1*2

问题太多.1,第四行doubleresult,product,case;中case不能作变量名,保留字.2,第十行result=fact(i);函数参数太少,你下面定义的函数应该有2个参数3,case

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

#include#includevoidmain(){intn=2;floatx;doublesum=1.0,term=1.0;printf("inputx:");scanf("%f",&x);do{

#includevoidmain(){intx,y;scanf("%f",&x);if(x=10){y=3*x-11;printf("%f",y);}else{y=2*x-1;printf("%f",