编写一函数计算∑_(x=1)^n▒X^k

#includevoidmain(){doublei,j;doublea,b,c=0;printf("请输入积分上下限:");scanf("%lf%lf",&a,&b);for(i=a;i

用系统内置的函数#include#includevoidmian(){doublepower,f;power=pow(x,y);f=power;printf("f(x,y)=%f",pow(x,y))

程序如下：#include#includeintmain(){doublex,y;printf("pleaseinputanumber:");scanf("%lf",&x);if(x>1){y=x*x

用递归函数#include#includeintinv(intx);main(){intsum=0,i;for(i=1;i

1 #include <stdio.h> 2#include<math.h> 3#definepi3.14159265358 

cal.m文件如下：function y = cal(n)    c  = 1:n; &n

intf(intx){returnx*x-3*x+5;}voidfun(intn){for(inti=1;i

是用C语言么?我用递归调用帮你写个#includevoidmain(){intfac(intn);inta;floatb;prinf("inputanintegernumber:")scanf(%d,

很高兴回答你的问题!#includeintfun(intn,intx){if(n==0)return1;intmul=x;for(inti=n;i>1;i--)mul*=x;returnfun(n-1

问题中每个分项的x的次数应该依次增加2,即1,3,5,7,9,.#includeusingnamespacestd;intmain(){doublepi=3.1415926;doublex,resul

#includevoidmain(){intx,y;//你用的是浮点型,我改成整形的了...你如果真的要用浮点的话...要加上小数点,如2.0等...printf("x=");scanf("%d",&

#include<IOSTREAM.H>#include<math.h>voidmain(){    doublex,y; &

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

#includeusingnamespacestd;voidjishu(intn){intk=1;inta=1;doubles=0;for(inti=0;i

ifx>=0theny=xelsey=-xendifprinty

#includedoublesum(doublex,intn){\x05doubles=1;\x05inti;\x05doublet=1;\x05doublea=1;\x05for(i=1;i再问：以

495=99*5445=89*5这个等式你把它写成直线左边'_'为x,右边'_'为y则可得到x=5y-36;这样就可用for判断了

#includeusingnamespacestd;voidmain(){inta=0,b=0;cin>>a>>b;cout

#includevoidmain(){inta,b,sum=0;printf("请输入两个整数：");//将两个改成n个就好了scanf("%d%d",&a,&b);sum=a+b;printf("%

没法运行不过while(fabs(term)>=1e-5)不好比较吧你两边乘以10000再比较试试再问：不好意思没明白你的意思我想问问我哪个地方出错了