已知double y=608576;则表达式(int)(y*10+0.5)/10.0的值为
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已知double y=608576;则表达式(int)(y*10+0.5)/10.0的值为
class Point{ private double x; private double y; Point(double x1,double y1){ x=x1; y=y1; } public
#includeusing namespace std;double tiaohe (double,double);int main(){double x,y;coutx>>y&&x*y!=0) {double t_tiaohe=tiaohe(x,y);cout
double pow(double x,double y) C语言#include#includeint main(){double pn[5],t[5],p0=10000;t[0]=double pow(double 1.50,double 2.0);pn[0]=p0*t[0];t[1]=double pow(double 4.10,double 8.0);pn[1]=p0*t[1];t[2]=double pow(double 4.30,double 4.0);pn[2]=p0*t[
c++习题 10.下列函数定义中正确的是( ).请解析一下每个选项.还有正确答案A.double fun(double &x,double &y) B.double fun(double &x,double &y){ {double z; return *x/*y;z=x/y; }return z;}C.double fun(double *x,double *y) D.double
已知有如下类的定义:public class Point { private double x,y; public Point(double a,double b) { x=a;y=b; }public double getX(){return x;}public double getY(){return y;}} 编写一个测试类 TestPoint,在此类中定义一个方法,此方法
double x=1.52,y=5.
java问题..谢谢lass Point { double x,y,z; Point(double _x,double _y,double _z) { x = _x; y = _y;class Point {double x,y,z;Point(double _x,double _y,double _z) {x = _x;y = _y;z = _z;}void setX(double _x) {x = _x;}double getDistance(Point p) {return
编程中各种符号的意思麻烦解释一下#include double x = 1.0,y = 2.0;double add( double x,double y ) {double sum;sum = x + y;}double average( double x,double y ) {double result;result = add( x,y ) / 2.0;return result;}int main( void) {doub
c++二分法求方程的解已知函数 y = e^x+ln(x)-1,实现函数long double solve(long double y){// here}对于传入的y,返回x值.要求误差小于1e-6.
函数指针可以指向返回值相同但参数个数不同的函数吗?#include double f1(double x){ return x*x; }double f2(double x,double y){ return x*y; }double fun(double a,double b){//f = f2 ; r2 = (*f)(a,b);return r1 + r2;}main(){ double x1=5
double ** p=new double*
double(*y)(double)在C++中是什么意思
帮忙看看这段程序哪错了,非常感激!#include #include double f(double x) { double y;#include#includedouble f(double x){double y;y=sin(x);y=y/x;return y;}void main(){double a,b,r,h,z,n;int k,j;double p[50][50];printf(Please input'a=?,b=?,r
c报错 invalid operands of types 'double' and 'double()(double)' to binary 'operator/'是怎么回事啊double mutual(double a,double b,double f){double y;y=(-(b*a+f*(1-a))*log(b*a+f*(1-a))-((1-b)*a+(1-f)*(1-a))*log((1-b)*a+(1-f)*(1-a))-a*(-b*log(b)
class Point{ public double x,y; Point(double x,double y){ this.x=x; this.y=y; } static void show(){老是出错,感激不尽class Point{public double x,y;Point(double x,double y){this.x=x;this.y=y;}static void show(){System.out.println(x=+x+ );Sy
以下函数用以求x的y次方 double fun(double x,int y){int i;double z=1.0;for(i=1;i
大大们写一下,注释啊,求class Point {double x,y,z;Point(double _x,double _y,double _z) {x = _x;y = _y;z = _z;}void setX(double _x) {x = _x;}double getDistance(Point p) {return (x - p.x)*(x - p.x) + (y - p.y)*(y - p.y) + (z - p.z)*(z - p.z);}}