编写函数double,计算两参数平方差的绝对值-搜答案-搜搜做题作业网

#includevoidmain(){doublex,y;{printf("请输入x=");}scanf("%lf",&x);if(x>=-1.0&&x

#include <iostream>using namespace std;double Max(double a, doub

#includemain(){doublex,y;scanf("%lf",&x);if(x

试试下面这个#includevoidmain(void){doublefunc(int);doubles=0,term;inti;for(i=0,term=1;;i++){term=1.

publicclassTest{　　publicstaticvoidmain(String[]args){　　doublen=10;　　doubleresult=fmethod(n);　　System

doublegetSquare(doublen){returnpi*n*n;}doublegetCircle(doublen){return2*pi*n;}

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

#include#includedoublefun(intn){doublesum=0.0;inti;intflag=-1;for(i=1;i{flag=(-1)*flag;sum+=1.0/i;}r

#includedoublefun(doubleh){returnfloor(h*100+0.5)/100;}

doublefun(doublex,doubley){doubler=x*x-y*y;returnr>0?r:-r;}再问：就这样就行了？？不用详细点吗？再答：double r=x*x-y*

#includeusingnamespacestd;doublefun(intn,doubleh){if(n==1)returnh;elseif(n再问：usingnamespacestd;这句

doublerect_area(doublelength,doublewidth=0){if(width==0.0)width=length;returnlength*width;}

将x打印成字符串,用'.'分割字符串,split[0]和[1]都转化成整数即可.注意字符串长度不要越界即可

#includelongfac(intn){inti;longx=1;for(i=2;i再问：谢谢咯！可是我说的是递归法哦！再答：#includelongfac(intn){if(n==0)retur

fun函数是double类型的,而返回的y是int类型,二者不匹配,将y强制为double类型就可以了

#include<math.h>double root(double x1,double x2){

#includeintmain(){intn;doublem;doublefac(intn);scanf("%d",&n);m=fac(n);printf("%d!=%f\n",n,m);return

double average(double score[], int n){ double sum=0.0;&nbs

#include#includedoublefun(intn)//计算1!+2!+3!+.+n!,并赋值给fun{inti;doubles=0,t=1;for(i=1;i

int main( ){ int s,e,i,n,A[100]; double m;&nbs