编写函数double,计算两参数平方差的绝对值
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#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