integral(float(*f)(float a,float

integral(float(*f)(float a,float c++ subscript is not of integral type#include #include using namespace std;int count(float x0,float f0,float i,int E,int N){ float x[20]; double f[20]; float fd[20]; float fd0=2*x0*x0-1; for(i=1;i 下列语句定义pf为指向float类型变量f的指针,__B___是正确的.B．float f,*pf = &f;下列语句定义pf为指向float类型变量f的指针,__B___是正确的.A．float f,*pf = f; B．float f,*pf = &f;C．float *pf = &f,f; D.float f,*pf float a1=2.0f,后面这个'f'表示意义? float f=123.456; printf(%10.2f,%-10.1f ,f,f);中的-10.1怎么解释 求x^3-15x^2+16x-80一个实根,下面程序怎么算不出来#include#includefloat f(float x){return ((x-5)*x-16)*x-80;}float xpoint(float x1,float x2){return (x1*f(x2)-x2*f(x1))/(f(x2)-f(x1));}float root(float x1,float x2){float x,y,y1;y1=f(x1);do{ integral float f1 = 2.2f; float f2 = 2.2; f1 f2 各等于什么 C语言：下面的功能是计算函数F（x,y,z）=（x+y）*（x+y）+（x+y）*（x-y）的值并输出.float f（float,float）；main（）{float x,y,z,sum；scanf（%f%f%f,&x&y&z）;sum=_____________;printf(sum=%f' ,sum);}float f(float a, #include main() { float x,y; scanf(%f,&x); if(x int main() { float a=1.0f; cout 解方程x *x *x - 5.0 *x *x + 16.0 *x - 80.0 下面是程序,谁给我讲讲原理啊#include #include float f(float x){float y;y = x *x *x - 5.0 *x *x + 16.0 *x - 80.0;return y; }float xpoint(float x1,float x2){float y;y = (x1 *f(x2) -x2 *f(x1))/(f 这个变量定义哪错了?float f=9.8; (9.8f)哪错了?是float f=9.8f；? 关于定积分~错哪儿了?急~# include # define N 100# define F(x) (x)*(x)*(x)float sab(float a,float b){ float h,sum=0; scanf(%f%f,&a,&b); h=(b-a)/N; for(;a c语言的问题,求x^3-5x^2+16x-80的根,我用的弦截法,但是出错#include#includefloat f(float x){float y;y=x*(x*x-5*x+16)-80;return y;}float xpoint(float x1,float x2){float root;root=(f(x2)*x1-x2*f(x1))/(f(x2)-f(x1));return root;}void main() 麻烦给讲讲这道题有以下程序float f1(float n){ return n*n; }float f2(float n){ return 2*n; }main(){ float (*p1)(float),(*p2)(float),(*t)(float),y1,y2;p1=f1; p2=f2;y1=p2( p1(2.0) );t = p1; p1=p2; p2 = t;y2=p2( p1(2.0) ); printf(%3.0f,%3.0f 编写由三角形三边求面积的函数.#includefloat kig(float x, float y,float z);main(){ float s,a,b,c; scanf(%f%f%f,&a,&b,&c); s=kig(a,b,c); printf(s=%f ,kig(a,b,c));}float kig(float x,float y,float z){ float p; p=1/4*(2*x*z+y*y-x*x-z*z); #include void main(){ float F; float c; scanf(%f,&F); c=5(F-32)/9; printf(%f,c); }显示的是error term does not evaluate to a function 为什么就是运行不出来?哪里粗错了?#include void main(){float F;float c;scanf(%f,&F);c=5(F-32)/