C语言编程习题.docx

1、C语言编程习题C语言编程习题第二章习题2-25. 用二分法编程求 6x4 -40x2+9=0 的所有实根。#include #include #define N 10000double A,B,C;double f(double x) return (A*x*x*x*x+B*x*x+C);void BM(double a,double b,double eps1,double eps2) int k; double x,xe; double valuea = f(a); double valueb = f(b); if (valuea 0 & valueb 0 | valuea 0 & val

2、ueb 0) return; printf(Finding root in the range: %.3lf, %.3lfn, a, b); for(k=1;k=N;k+) x=(a+b)/2; xe=(b-a)/2; if(fabs(xe)eps2 | fabs(f(x)eps1) printf(The x value is:%gn,x); printf(f(x)=%gnn,f(x); return; if(f(a)*f(x)0) b=x; else a=x; printf(No convergence!n);int main() double a,b,eps1,eps2,step,star

3、t; printf(Please input A,B,C:n); scanf(%lf %lf %lf,&A,&B,&C); printf(Please input a,b, step, eps1,eps2:n); scanf(%lf %lf %lf %lf %lf,&a,&b,&step,&eps1,&eps2); for (start=a; (start+step) = b; start += step) double left = start; double right = start + step; BM(left, right, eps1, eps2); return 0;运行：Ple

4、ase input A,B,C:6 -40 9Please input a,b, step, eps1,eps2:-10 10 1 1e-5 1e-5Finding root in the range: -3.000, -2.000The x value is:-2.53643f(x)=-0.00124902Finding root in the range: -1.000, 0.000The x value is:-0.482857f(x)=0.00012967Finding root in the range: 0.000, 1.000The x value is:0.482857f(x)

5、=0.00012967Finding root in the range: 2.000, 3.000The x value is:2.53643f(x)=-0.00124902有时若把判别语句if(fabs(xe)eps2 | fabs(f(x)eps1)改为if(fabs(xe)eps2 & fabs(f(x)eps1)会提高精度，对同一题运行结果：Finding root in the range: -3.000, -2.000The x value is:-2.53644f(x)=-4.26496e-007Finding root in the range: -1.000, 0.000T

6、he x value is:-0.482861f(x)=-7.3797e-006Finding root in the range: 0.000, 1.000The x value is:0.482861f(x)=-7.3797e-006Finding root in the range: 2.000, 3.000The x value is:2.53644f(x)=-4.26496e-007习题2-35. 请用埃特金方法编程求出x=tgx在4.5（弧度）附近的根。#include #include #define N 100#define PI 3.1415926void SM(double

7、 x0,double eps) int k; double x; double x1,x2; for(k=1;k=N;k+) x1=sin(x0)/cos(x0); x2=sin(x1)/cos(x1); x=x0-(x1-x0)*(x1-x0)/(x2-2*x1+x0); if(fabs(x-x0)eps) printf(The x value is:%gn,x); return; x0=x; printf(No convegence!n);int main() double eps,x0; printf(Please input eps,x0:n); scanf(%lf %lf,&eps,

8、&x0); SM(x0,eps); return 0;运行:Please input eps,x0:1e-5 4.5The x value is:4.49341习题2-411请编出用牛顿法求复根的程序，并求出 P(z)=z4-3z3+20z2+44z+54=0 接近于z0=2.5+4.5i 的零点。#include #include #define MAX_TIMES 1000typedef struct double real, image; COMPLEX;COMPLEX Aa5=54,0,44,0,20,0,-3,0,1,0;COMPLEX Bb4=44,0,40,0,-9,0,4,0;

9、COMPLEX zero = 0, 0;double eps1=1e-6;double eps2=1e-6;COMPLEX multi(COMPLEX a,COMPLEX b) COMPLEX result; result.real = a.real * b.real - a.image * b.image; result.image = a.image * b.real + a.real * b.image; return result;COMPLEX Div(COMPLEX a,COMPLEX b) COMPLEX z3; double s; s=(b.real*b.real)+(b.im

10、age*b.image); z3.real=b.real; z3.image=-b.image; z3=multi(a,z3); z3.real=z3.real/s; z3.image=z3.image/s; return z3;COMPLEX add(COMPLEX a,COMPLEX b) COMPLEX result; result.real = a.real + b.real; result.image = a.image + b.image; return result;COMPLEX subtract(COMPLEX a, COMPLEX b) COMPLEX result; re

11、sult.real = a.real - b.real; result.image = a.image - b.image; return result;COMPLEX times(COMPLEX z,int n) int i; COMPLEX result=1, 0; for (i=0;in;i+) result=multi(result,z); return result;double distance(COMPLEX a, COMPLEX b) return sqrt(a.real - b.real) * (a.real - b.real) + (a.image - b.image) *

12、 (a.image - b.image);double complex_abs(COMPLEX a) return sqrt(a.real * a.real + a.image * a.image);COMPLEX f(COMPLEX x) int i; COMPLEX result=zero; for (i=0;i5;i+) result=add(result,multi(Aai,times(x,i); return result;COMPLEX ff(COMPLEX x) int i; COMPLEX result=zero; for (i=0;i4;i+) result=add(resu

13、lt,multi(Bbi,times(x,i); return result;int main() COMPLEX z0,z1,result; double x,y; int k; printf(please input x,yn); scanf(%lf %lf,&x,&y); z0.real=x; z0.image=y; for(k=0; kMAX_TIMES; k+) z1 = subtract(z0, Div(f(z0), ff(z0); result = f(z1); if (distance(z0,z1)eps1 | complex_abs(result)eps2) printf(T

14、he root is: z=(%.3f + %.3fi), f(z)=(%e + %ei)n, z1.real,z1.image, result.real, result.image); return 0; z0 = z1; printf(No convergence!n); return 0;运行:please input x,y2.5 4.5The root is: z=(2.471 + 4.641i), f(z)=(-1.122705e-007 + -1.910245e-007i)习题2-62. 请编程用劈因子法求高次方程 x4+ x 3+5 x 2+4 x +4=0的所有复根。#inc

15、lude#includeint main() float b20,c20; float delta; float eps; int print=0; float fru0,fru1,s0,s1,frv0,frv1; float deltau,deltav; float u,v; float a20; int n,j; float r0,r1; float r; int i, k; printf(Please input max exponent:n); scanf(%d,&n); printf(Please input epslion:n); scanf(%f,&eps); printf(Pl

16、ease input the coefficient:n); for(i=0;i=n;i+) scanf(%f,&ai); for(j=0;j=n;j+) printf(+%.3fX%d,aj,n-j); printf(n); for(k=1;k=2;k+) u=an-1/an-2; v=an/an-2; if(k=2) u=0;v=4; while(1) b0=a0; b1=a1-u*b0; for(j=2;j=n;j+) bj=aj-u*bj-1-v*bj-2; r0=bn-1; r1=bn+u*bn-1; c0=b0; c1=b1-u*b0; for(j=2;j=n-2;j+) cj=b

17、j-u*cj-1-v*cj-2; s0=cn-3; s1=cn-2+u*cn-3; fru0=u*s0-s1; fru1=v*s0; frv0=-s0; frv1=-s1; deltau=-(frv1*r0-frv0*r1)/(fru0*frv1-fru1*frv0); deltav=-(fru1*r0-fru0*r1)/(frv0*fru1-frv1*fru0); u=u+deltau; v+=deltav; delta=u*u-4*v; if(fabs(deltau)eps)&(fabs(deltav)eps) print=1; printf(found roots:); r=-u/2;

18、i=(int) sqrt(fabs(delta)/2; if(delta0) printf(%.3f+%.3fit,r,fabs(double)i); printf(%.3f-%.3fin,r,fabs(double)i); else printf(%.3ftt,r+i); printf(%.3fn,r-i); if(n=1) printf(Found root :%.3fn,a1/a0); if(k=2) return 0; if(k=1&print) break; /* end while */ /* end for */ return 0;运行:Please input max expo

19、nent:4Please input epslion:1e-5Please input the coefficient:1 1 5 4 4+1.000X4+1.000X3+5.000X2+4.000X1+4.000X0found roots:-0.500+0.000i -0.500-0.000ifound roots:0.000+2.000i 0.000-2.000i习题3-15. 请编程用列全主元高斯约当消去法求矩阵A, B的逆矩阵。 #include #include #define MAX 10int ROW;static float AMAX+12*MAX+1;void putout(

20、) int i,j; printf(nThe reverse matrix is:n); for(i=1;i=ROW;i+) for(j=ROW+1;j=2*ROW;j+) printf(%.3f ,Aij); printf(n); int get() float maxMAX+1; int count_rMAX+1; float tmp2*MAX+1; float pass1,pass2; int i,j,k; for(i=1;i=ROW;i+) maxi=Aii; count_ri=i; for(k=i;k=ROW;k+) if(maxiAki) maxi=Aki; count_ri=k;

21、 /* find the max one*/ if(maxi=0) return -1; if(count_ri!=i) for(k=1;k=2*ROW;k+) tmpk=Acount_rik; Acount_rik=Aik; Aik=tmpk; /*change the rows*/ pass1=Aii; for(k=1;k=2*ROW;k+) Aik/=pass1; for(j=1;j=ROW;j+) if(j=i) continue; else pass2=Aji; for(k=1;k=MAX) printf(the number of row is too big!n); return

22、 0; printf(Please input the matrix A:n); for(i=1;i=ROW;i+) for(j=1;j=ROW;j+) scanf(%f,&p); Aij=p; if(tmpfabs(Aij) tmp=fabs(Aij); if(tmp=0) printf(No inverse!); return 0; for(i=1;i=ROW;i+) Aii+ROW=1; return get();运行：please input the number of ROW:3Please input the matrix A:-3 8 52 -7 41 9 -6The rever

23、se matrix is:0.026 0.396 0.285 0.068 0.055 0.094 0.106 0.149 0.021please input the number of ROW:4Please input the matrix A:2 1 -3 -13 1 0 7-1 2 4 -21 0 -1 5The reverse matrix is:-0.047 0.588 -0.271 -0.9410.388 -0.353 0.482 0.765-0.224 0.294 -0.035 -0.471-0.035 -0.059 0.047 0.294习题3-24. 编程用追赶法解下列三对角

24、线方程组。#include #include #define ROW 5static float AROW+1=0,0,0,0,0,0;static float BROW+1=0,0,0,0,0,0;static float CROW+1=0,0,0,0,0,0;static float XROW+1=0,0,0,0,0,0;int main() int i; printf(Please input the A:n); for(i=2;i=ROW;i+) scanf(%f,&Ai); printf(Please input the B:n); for(i=1;i=ROW;i+) scanf(%

25、f,&Bi); printf(Please input the C:n); for(i=1;i=ROW-1;i+) scanf(%f,&Ci); printf(Please input the F:n); for(i=1;i=ROW;i+) scanf(%f,&Xi); C1=C1/B1; for(i=2;i=ROW-1;i+) Ci=Ci/(Bi-Ai*Ci-1); X1=X1/B1; for (i=2; i=1;i-) Xi=Xi-Ci*Xi+1; printf (The value is : n); for(i=1;i=ROW;i+) printf (x%d=%.3fn,i,Xi); return 0;运行：Please input the A:-1 -1 -1 -1Please input the B:4 4 4 4 4 Please input the C:-1 -1 -1 -1Please input the F:100 200 200 200 100The value is : x1=46.154x2=84.615x3=92.308x4=84.615x5=46.154习题4-31. 用高斯赛