用C解决一元二次方程根的问题集锦.docx

1、用C解决一元二次方程根的问题集锦c+求一元二次方程ax+bx+c=0的根一#include #include #define ESSP 0.00000001 void main() double a,b,c,delta,root1,root2; couta; coutb; coutc; delta=b*b-4*a*c; if(delta-ESSP) coutThis equation has no real root.ESSP) root1=(-b+sqrt(delta)/(2*a); root2=(-b-sqrt(delta)/(2*a); coutThis eqution has two

2、unequalize roots:root1root2endl; else root1=-b/(2*a); coutThis eqution has two equalize roots:root1endl; 二结构化编程方法: /for two roots #include #include int main() double a,b,c,rp,ip;/方程系数 double x1,x2;/方程的根 double d;/根的判别式 coutabc; if(a=0) cout= 0) if (d=0) x1 = -b / (2*a); cout只有一个实根:x1endl; else x1 =

3、( -b + sqrt ( d ) ) / ( 2 * a ); x2 = ( -b - sqrt ( d ) ) / ( 2 * a ); cout有二个实根:x1andx2endl; else rp = -d/(2*a); ip = sqrt ( -d ) /(2*a); x1 = rp + ip ; x2 = rp - ip ; cout有二个虚根:x1iandx2i0) x1=(-b+sqrt(d)/(2*a); x2=(-b-sqrt(d)/(a*a); else if(d=0) x1=x2=(-b)/(2*a); else double rp = -d/(2*a); double

4、ip = sqrt ( -d ) /(2*a); x1=rp + ip; x2=rp - ip; void FindRoot:Display() if(d0) coutx1=x1nx2=x2endl; else if(d=0) coutx1=x2=x1endl; else coutx1=x1iendl; coutx2=x2iendl; void main() FindRoot obj(1,2,3); obj.Find(); obj.Display(); 三根据要求编写程序。 从键盘输入三个实数a、b、c分别作为一个一元二次方程ax2+bx+c=0 的三个系数。使用系 统给出的平方根函数，编写一

5、段程序，使之求出这个方程的两个根。其中，求=b2-4*a*c 的功能要以函数形式出现。（提示：求根公式，0) x1=(-b)+sqrt(disc); x2=(-b)-sqrt(disc); printf(x1=%.2ftx2=%.2fn,x1,x2); else printf(b-4ac0,该方程没有实根n,x1,x2); #include stdio.h int a8=0; int getResult(int n) if(n=1|n=2) return 1; else return 3*getResult(n-1)-getResult(n-2); void main() int i; a0=

6、1; a1=1; for(i=2;i8;i+) ai=getResult(i+1); for(i=0;i8;i+) printf(a%d = %dn,i+1,ai); 四用C语言编写一个程序，计算一元二次方程的解(x1，x2)。一元二次方程：Ax2+Bx+C=0 (假设A=4,B=6,C=1)按顺序输入a,b,c #include stdio.h #include stdlib.h #include math.h int main() float a,b,c; scanf(%f,&a); scanf(%f,&b); scanf(%f,&c); printf(%fn,(-b+sqrt(b*b-4

7、*a*c)/2/a); printf(%f,(-b-sqrt(b*b-4*a*c)/2/a); system(pause); return 0; ;五解一元二次方程的C+程序 #include #include #include void sqrt1(double,double); void sqrt2(double,double); void printer(double,double); void main() float a,b,c; double x1,x2; cout请输入a，b，c：abc; if(b*b-4*a*c !=0) if(b*b-4*a*c) sqrt1(x1,x2);

8、 else printer(x1,x2); else sqrt2(x1,x2); void sqrt1(double x1,double x2) /b*b-4ac0 float a,b,c; if(a!=0) x1=(-b+sqrt(b*b-4*a*c)/(2*a); x2=(-b-sqrt(b*b-4*a*c)/(2*a); coutx1=x1setw(5)x2=x2endl; else if(b=0) if(c=0) coutx为任意解endl; else cout无解endl; else x1=x2=-c/b; coutx1=x2=x1endl; void sqrt2(double x1

9、,double x2) /b*b-4ac=0 float a,b,c; x1=x2=-b/(2*a); coutx1=x2=x1endl; void printer (double x1,double x2) /b*b-4ac0 float a,b,c; char i=i;/i后面为虚部 x1=(-b/(2*a)+(i*(sqrt(b*b-4*a*c)/(2*a); x2=(-b/(2*a)-(i*(sqrt(b*b-4*a*c)/(2*a); coutx1=x1tx2=x20，有两个不等实根。 (4) b2-4ac0，有两个复根（表示成x+yi,x-yi）。 2编写一个程序，打印下列图形：

10、1 1 2 1 2 3 1 2 3 4 1 2 3 4 5 3. 编写一个函数将一nxn矩阵转置（即aij与aji的值交换）,然后编写一main( ) 函数调用它（取n=3）。第1个我记得书上有例题的 第2个： #include int main() for(int i=1;i=5;i+) for(int j=1;j=i;j+) printf(%d,j); printf(n); return 0; 3. #include #define N 3 int main() void f(int); int aNN; for(int i=0;iN;i+) for(int j=0;jN;j+) scan

11、f(%d,&aij); f(a); for(int i=0;iN;i+) for(int j=0;jN;j+) printf(%d,aij); printf(n); return 0; void f(int a) int temp; for(int i=0;iN;i+) for(int j=i;jN;j+) temp=aij; aij=aji; aji=temp; 七（复数）#include iostream.h #include math.h void main() float a,b,c; couta; coutb; coutc; if(b*b-4*a*c0) coutNo radix i

12、n real number!endl; complex x1,x2; x1 = complex(-b)/(2*a), sqrt(-d)/(2*a); x2 = complex(-b)/(2*a), -(sqrt(-d)/(2*a); return; 八 建立复数类求方程的根#include #include #include using namespace std; int main(void) double a = 1.0, b = 0.0, c = 1.0; / ax2 + bx + c = 0 complex x1,x2; double d = b * b - 4.0 * a * c;

13、if (d = 0) x1 = (-b + sqrt(d)/(2*a); x2 = (-b - sqrt(d)/(2*a); else x1 = complex(-b)/(2*a), sqrt(-d)/(2*a); x2 = complex(-b)/(2*a), -(sqrt(-d)/(2*a); cout x1 = x1.real() + x1.imag() i endl; cout x2 = x2.real() + x2.imag() i endl; return 0; 九复数类如下: #include #include / class class complex private: dou

14、ble a,b; double _copy; public: double getreal(); double getimag(); double getZ(); void solution(double a,double b,double c,double p,complex &x1,complex &x2); complex(int x) a=x;b=0; complex(double av,double bv):a(av),b(bv) complex() complex(complex& _copy) a=_copy.a; b=_copy.b; friend ostream& opera

15、tor(istream& g,complex& i); friend complex operator+(complex& x,complex& y); friend complex operator-(complex& x,complex& y); friend complex operator*(complex& x,complex& y); friend complex operator/(complex& x,complex& y); complex operator=(complex& y); void output() coutz=sqrt(a*a)+(b*b)endl; void

16、 put() cout(a+bi)endl; void gen() int sign=1, double g,h; if(b0) sign=-1; else sign=1; g=sqrt(sqrt(a*a)+(b*b)+a)/2); h=sign*sqrt(sqrt(a*a)+(b*b)-a)/2); coutthe sqrt-root is: g+hiendl; ; ostream& operator(istream& stream,complex& x) coutx.a; coutx.b; return stream; /add complex operator+(complex& x,c

17、omplex& y) return complex(x.a+y.a,x.b+y.b); /sub complex operator-(complex& x,complex& y) return complex(x.a-y.a,x.b-y.b); /mul complex operator*(complex& x,complex &y) return complex(x.a*y.a)-(x.b*y.b),(x.a*y.b)+(x.b*y.a); /div complex operator/(complex& x,complex& y) return complex(x.a*y.a+x.b*y.b)/(y.a*y.a+y.b*y.b),(x.b*y.a-x.a*y.b)/(y.a*y.a+y.b*y.b); /assignment complex complex:operator=(complex& y) this-a=y.a; this-b=y.b; return *this; double complex:getreal() return a; double complex:getimag() return b; double complex:getZ(