C语言解题经典程序(2)

totoerer · 发表于 2004-5-14 08:54:30

<

center" align=center>/* 矩阵乘法运算（用指针） AMUL1.C */<

18pt">#include<stdio.h><

18pt">void mul_array(double *a,double *b,double *c,int m,int s,int n) { int i,j,k; for(i=0;i<m;i++) for(j=0;j<n;j++) { *(c+i*n+j)=0.0; for(k=0;k<s;k++) { *(c+i*n+j)+=*(a+i*s+k)**(b+k*n+j); } } } main() { int i,j,m=5,s=3,n=4; double a[5][3]={3,1,6,-2,2,5,7,4,-1,5,-3,2,-3,4,3}; double b[3][4]={-2,2,1,3,3,-1,2,4,4,-2,1,-5}; double c[5][4]; mul_array(a,b,c,m,s,n); for(i=0;i<m;i++) { printf("\n"); for(j=0;j<n;j++) printf("c[%d][%d] = %g ",i,j,c[j]); } }

totoerer · 发表于 2004-5-14 08:55:20

<

center" align=center>矩阵乘法运算(用二级指针)amul3.c<

22pt; mso-line-height-rule: exactly">#include<stdio.h><

22pt; mso-line-height-rule: exactly">void mul_array(double **a,double **b,double **c,int m,int s,int n) { int i,j,k; for(i=0;i<m;i++) for(j=0;j<n;j++) { c[j]=0.0; for(k=0;k<s;k++) c[j]+=(a[k]*b[k][j]); } }main() { int i,j,m=5,s=3,n=4; double c[5][4]; double *pa[5],*pb[3],*pc[5];double a[5][3]={3,1,6,-2,2,5,7,4,-1,5,-3,2,-3,4,3}; double b[3][4]={-2,2,1,3,3,-1,2,4,4,-2,1,-5}; for(i=0;i<5;i++) { pa=a; pc=c; } for(i=0;i<3;i++)pb=b; mul_array(pa,pb,pc,m,s,n); for(i=0;i<m;i++) { printf("\n"); for(j=0;j<n;j++) printf("c[%d][%d] = %g ",i,j,c[j]); } }

totoerer · 发表于 2004-5-14 08:55:48

<

center" align=center>矩阵乘法运算（用指针）amul1.c<

23pt; mso-line-height-rule: exactly">#include<stdio.h><

23pt; mso-line-height-rule: exactly">void mul_array(double *a,double *b,double *c,int m,int s,int n) { int i,j,k; for(i=0;i<m;i++) for(j=0;j<n;j++) { *(c+i*n+j)=0.0; for(k=0;k<s;k++) *(c+i*n+j)+=*(a+i*s+k)**(b+k*n+j); } }main() { int i,j,m=5,s=3,n=4; double c[5][4];double a[5][3]={3,1,6,-2,2,5,7,4,-1,5,-3,2,-3,4,3}; double b[3][4]={-2,2,1,3,3,-1,2,4,4,-2,1,-5}; mul_array(a,b,c,m,s,n); for(i=0;i<m;i++) { printf("\n"); for(j=0;j<n;j++) printf("c[%d][%d] = %g ",i,j,c[j]); } }

totoerer · 发表于 2004-5-14 08:58:55

<

center" align=center>柯特斯公式计算定积分 Cotes1.C<

22pt; mso-line-height-rule: exactly">#include<math.h><

22pt; mso-line-height-rule: exactly">#include<stdio.h>double f(double x) { return(4.0/(1+x*x)); }double cotes(double a,double b,double eps) { int k=0; double h,h1,m1,m2,m3,m4,c1,c2,x,rc; h=0.25*(b-a); h1=h+h; rc=2.0/45.0; m1=(f(a)+f(b))*7.0; m2=f(a+h)+f(b-h); m3=f(a+h1); c1=(m1+m2*32.0+m3*12.0)*h*rc; m4=0.0; while(k<=20) { h1=h;h*=0.5;m4+=m3;m3=m2; m2=0.0;x=a+h; while(x<b) { m2+=f(x);x+=h1; } c2=(m1+m2*32.0+m3*12.0+m4*14.0)*h*rc; if(fabs(c2-c1)<eps)break; c1=c2; k++; } printf("\nk=%d",k); return(c2); }main( ) { double eps=1.0e-12,answer; answer=cotes(0.0,1.0,eps); printf("\nThe Answer is %15.14lf ",answer); } /* 循环5次计算结果为 3.14159265358979 */

totoerer · 发表于 2004-5-15 07:05:26

<

center" align=center>拉格朗日插值法通用计算程序LACA.C<

24pt; mso-line-height-rule: exactly">#include<stdio.h><

24pt; mso-line-height-rule: exactly">#include<conio.h>double laca(double xi[],double yi[],int n,double x){int k,j; double p,y=0.0;for(k=0;k<=n;k++) { p=1.0; for(j=0;j<=n;j++) if(k!=j)p=p*(x-xi[j])/(xi[k]-xi[j]); y=y+p*yi[k]; }return(y);}main(){double xi[]={64.0,81.0,100.0,121.0,144.0,169.0};double yi[]={8.0,9.0,10.0,11.0,12.0,13.0};double x＝115.0,y;y=laca(xi,yi,5,x);printf("\nx=%g y=%10.8lf",x,y);}

totoerer · 发表于 2004-5-15 07:06:26

<

center" align=center>离散牛顿法解非线性方程组 lisannt.c<

23pt; mso-line-height-rule: exactly">#include<math.h><

23pt; mso-line-height-rule: exactly">#include<stdio.h>#define N 3double a[N][N+1];double pi;double f1(double x[]){return(3.0*x[0]-cos(x[1]*x[2])-0.5);}double f2(double x[]){return(x[0]*x[0]-81.0*(x[1]+0.1)*(x[1]+0.1)+sin(x[2])+1.06);}double f3(double x[]){return(exp(-x[0]*x[1])+20.0*x[2]+pi*10.0/3.0-1.0);}void FDF(double x[],double h){int i;double y1,y2,y3,x1[N];y1=f1(x);y2=f2(x);y3=f3(x);a[0][N]=y1;a[1][N]=y2;a[2][N]=y3;for(i=0;i<N;i++)x1=x;x1[0]+=h; a[0][0]=(f1(x1)-y1)/h;a[1][0]=(f2(x1)-y2)/h; a[2][0]=(f3(x1)-y3)/h;x1[0]=x[0];x1[1]+=h; a[0][1]=(f1(x1)-y1)/h;a[1][1]=(f2(x1)-y2)/h; a[2][1]=(f3(x1)-y3)/h;x1[1]=x[1];x1[2]+=h; a[0][2]=(f1(x1)-y1)/h;a[1][2]=(f2(x1)-y2)/h; a[2][2]=(f3(x1)-y3)/h;} int gauss(int n) { int i,j,k,i0; double c,t,eps=1.0e-20; for(k=0;k<n;k++) { c=0.0; for(i=k;i<n;i++) { t=fabs(a[k]); if(t>c) { c=t;i0=i; } } if(c<=eps) return(0); if(i0!=k) for(j=k;j<=n;j++) { t=a[k][j]; a[k][j]=a[i0][j]; a[i0][j]=t; } c=1.0/a[k][k]; for(j=k+1;j<=n;j++)a[k][j]*=c; for(i=k+1;i<n;i++) for(j=k+1;j<=n;j++) a[j]-=(a[k]*a[k][j]); } for(i=n-1;i>=0;i--) for(j=i+1;j<n;j++) a[n]-=(a[j]*a[j][n]); return(1);} int lisanfa(double x0[],double h,double eps){int i,k=1;double dx,x1[N],y[N];for(;;) { FDF(x0,h); if(!gauss(N))return(0); for(i=0;i<N;i++)y=a[N]; dx=0.0; for(i=0;i<N;i++)if(fabs(y)>dx)dx=fabs(y); for(i=0;i<N;i++) { x1=x0-y; x0=x1; } printf("\nk=%d",k); printf("\nx1=%12.10lf x2=%12.10lf x3=%12.10lf",x1[0],x1[1],x1[2]); if(dx<eps)break; h*=dx; if(fabs(h)<eps)h=eps; k++;if(k>200)return(0); }return(1);} main(){int i;double x0[N]={2.0,1.0,0.0},h=0.1,eps=1.0e-08;pi=4.0*atan(1.0);if(lisanfa(x0,h,eps)) { for(i=0;i<N;i++) printf("\nx[%d] = %14.12lf ",i,x0); }else printf("\nFiled !");}

totoerer · 发表于 2004-5-15 07:07:06

<

20pt">二、已知<v:shapetype> <v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0"></v:f><v:f eqn="sum @0 1 0"></v:f><v:f eqn="sum 0 0 @1"></v:f><v:f eqn="prod @2 1 2"></v:f><v:f eqn="prod @3 21600 pixelWidth"></v:f><v:f eqn="prod @3 21600 pixelHeight"></v:f><v:f eqn="sum @0 0 1"></v:f><v:f eqn="prod @6 1 2"></v:f><v:f eqn="prod @7 21600 pixelWidth"></v:f><v:f eqn="sum @8 21600 0"></v:f><v:f eqn="prod @7 21600 pixelHeight"></v:f><v:f eqn="sum @10 21600 0"></v:f></v:formulas><v:path gradientshapeok="t" connecttype="rect" extrusionok="f"></v:path><lock v:ext="edit" aspectratio="t"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape>展开成幂级数为<

12pt; LINE-HEIGHT: 20pt"> <v:shape><v:imagedata></v:imagedata></v:shape><

20pt">试编写程序，从键盘输入一个实数x，利用上述公式计算并输出<v:shape> <v:imagedata></v:imagedata></v:shape>的数值，计算误差控制<v:shape> <v:imagedata></v:imagedata></v:shape>。#include<stdio.h>main() { double n=1.0, e, x, t=1, xn, y; printf(“\nInput a number”); scanf(“%lf”,&x); e=1.0+x;xn=x;do { n+=1.0; t*=n; xn*=x; y=xn/t; e+=y; }while(y>1.0e-10); printf(“\nex=%12.10lf”,e);}输出结果：当x=1时，ex=2.71828182846

totoerer · 发表于 2004-5-15 07:07:56

<

24pt; mso-line-height-rule: exactly">#include<stdio.h><

24pt; mso-line-height-rule: exactly">main()<

24pt; mso-line-height-rule: exactly"> { float x,y;char op; printf("\n请输入两个数的运算表达式\n"); scanf("%f%c%f",&x,&op,&y); switch(op) { case '+':printf("结果是%f",x+y);break; case '-':printf("结果是%f",x-y);break; case '*':printf("结果是%f",x*y);break; case '/':printf("结果是%f",x/y);break; default:printf("输入有错误");break; } }说明：1．switch后面括号内的表达式应是整型或字符型数据。2．每个case后的常量的值互不相等。3．各case的先后次序无关紧要。4．若表达式的值与某个case后的常量相等，则执行此case后的语句，直到break语句为止，若找不到匹配，就执行default后的语句。5．每个case后的break语句不能省略。

totoerer · 发表于 2004-5-15 07:09:23

<

0cm 0cm 0pt; TEXT-INDENT: 0cm">/* 列主元法求逆矩阵 cainv.c */<

21pt; mso-line-height-rule: exactly">#include<math.h><

21pt; mso-line-height-rule: exactly">#include<stdio.h>int inverta(double *a[],double *b[],int n,double eps) { int i,j,k,i0; double c,t,maxa; for(i=0;i<n;i++) for(j=0;j<n;j++) if(j==i)b[j]=1.0; else b[j]=0.0; for(k=0;k<n;k++) { maxa=0.0; for(i=k;i<n;i++) { t=fabs(a[k]); if(t>maxa) { maxa=t;i0=i; } } if(maxa<=eps)return(0); if(i0!=k) { for(j=0;j<n;j++) { t=a[k][j]; c=b[k][j]; a[k][j]=a[i0][j]; b[k][j]=b[i0][j]; a[i0][j]=t; b[i0][j]=c; } } c=1.0/a[k][k]; for(j=k;j<n;j++)a[k][j]*=c; for(j=0;j<n;j++)b[k][j]*=c; for(i=k+1;i<n;i++) { for(j=k+1;j<n;j++) a[j]-=(a[k]*a[k][j]); for(j=0;j<n;j++) b[j]-=(a[k]*b[k][j]); } } for(k=n-2;k>=0;k--) for(i=k;i>=0;i--) for(j=0;j<n;j++)b[j]-=(b[k+1][j]*a[k+1]); return(1); }main() { int i,j,n=4; double b[4][4]; double eps=1.0e-9, *pa[4],*pb[4]; double a[4][4]={1,1,0,1,1,2,2,2,2,-2,2,1,3,-1,5,3}; for(i=0;i<n;i++) { pa=a; pb=b; } if(inverta(pa,pb,n,eps)) { for(i=0;i<n;i++) { printf("\n"); for(j=0;j<n;j++) printf("a-1[%d][%d]=%g ",i,j,b[j]); } } else printf("\nFiled !"); }

totoerer · 发表于 2004-5-15 07:10:35

<

center" align=center>列主元消去法解线性方程组 gauss.c<

23pt; mso-line-height-rule: exactly">#include<math.h><

23pt; mso-line-height-rule: exactly">#include<stdio.h>int gauss(double *a[],int n) { int i,j,k,i0; double c,t,eps=1.0e-20; for(k=0;k<n;k++) { c=0.0; for(i=k;i<n;i++) { t=fabs(a[k]); if(t>c) { c=t; i0=i; } } if(c<=eps)return(0); if(i0!=k) for(j=k;j<=n;j++) { t=a[k][j]; a[k][j]=a[i0][j]; a[i0][j]=t; } c=1.0/a[k][k]; for(j=k+1;j<=n;j++)a[k][j]*=c; for(i=k+1;i<n;i++)for(j=k+1;j<=n;j++) a[j]-=(a[k]*a[k][j]); } for(i=n-1;i>=0;i--) for(j=i+1;j<n;j++) a[n]-=(a[j]*a[j][n]); return(1);} main() { int i,j,n=3; double *pa[3]; double a[3][4]={1,1,1,6,2,3,1,11,1,2,3,14}; for(i=0;i<n;i++)pa=a; if(gauss(pa,n)) { for(i=0;i<n;i++) printf("\nx%d = %g ",i,a[n]); } else printf("\nFiled !"); }

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