C语言解题经典程序(2)

totoerer · 发表于 2004-5-15 07:37:09

<

center" align=center>/* 全主元消去法计算行列式的值 det.c */<

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

23pt; mso-line-height-rule: exactly">#include<stdio.h>double det(double **a,int n) { int i,j,k,i0,j0,m=1; double c,t,maxa,d=1.0,eps=1.0e-20; for(k=0;k<n-1;k++) { maxa=0.0; for(i=k;i<n;i++) for(j=k;j<n;j++) { t=fabs(a[j]); if(t>maxa) { maxa=t; i0=i; j0=j; } } if(maxa<=eps)return(0.0); if(i0!=k) { m=-m; for(j=k;j<n;j++) { t=a[k][j]; a[k][j]=a[i0][j]; a[i0][j]=t; } } if(j0!=k) { m=-m; for(i=k;i<n;i++) { t=a[j0]; a[j0]=a[k]; a[k]=t; } } d*=a[k][k]; c=1.0/a[k][k]; for(i=k+1;i<n;i++) { t=a[k]*c; for(j=k+1;j<n;j++) a[j]-=(a[k][j]*t); } } d=d*(double)m*a[n-1][n-1]; return(d); } main() { int i,n=4; double *pa[4]; double a[4][4]={1,2,3,4,2,3,4,1,3,4,1,2,4,1,2,3}; for(i=0;i<n;i++)pa=a; printf("\nD=%g",det(pa,n)); }

totoerer · 发表于 2004-5-15 07:41:20

<

center" align=center>/*全主元消去法求行列式的值*/<

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

18pt">#include<stdio.h> double det(double *a[],int n) { int i,j,k,i0,j0,m=1; double c,t,maxa,d=1.0,eps=1.0e-20; for(k=0;k<n-1;k++) { maxa=0.0; for(i=k;i<n;i++) for(j=k;j<n;j++) { t=fabs(a[j]); if(t>maxa) { maxa=t;i0=i;j0=j; } } if(maxa<=eps)return(0.0); if(i0!=k) { m=-m; for(j=k;j<n;j++) { t=a[k][j]; a[k][j]=a[i0][j]; a[i0][j]=t; } } if(j0!=k) { m=-m; for(i=k;i<n;i++) { t=a[j0]; a[j0]=a[k]; a[k]=t; } } d*=a[k][k]; c=1.0/a[k][k]; for(i=k+1;i<n;i++) { t=a[k]*c; for(j=k+1;j<n;j++) a[j]-=(a[k][j]*t); } } d=d*(double)m*a[n-1][n-1]; return(d); } main() { int i,n=4; double *pa[4]; double a[4][4]={1,2,3,4,2,3,4,1,3,4,1,2,4,1,2,3}; for(i=0;i<n;i++)pa=a; printf("\nD=%g",det(pa,n)); }

totoerer · 发表于 2004-5-15 07:41:45

<

center" align=center>赛德尔迭代法解线性方程组 seidel.c<

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

23pt; mso-line-height-rule: exactly">#include<stdio.h> int seidel(double *a[],double b[],double x[],double eps,int n){int i,j,k=1;double s,t,dx,maxdx;for(;;) { maxdx=0.0; for(i=0;i<n;i++) { t=x; s=b; for(j=0;j<n;j++) s+=(a[j]*x[j]); x=s; dx=fabs(s-t); if(dx>maxdx)maxdx=dx; } k++; printf("\nk=%d ",k); if(k>200)return(0); if(maxdx<eps)break; }return(1);}main(){int i,n=3;double a[3][3]={0,0.2,0.1,0.2,0,0.1,0.2,0.4,0};double x[3]={1,1,1},b[3]={0.3,1.5,2.0},eps=1.0e-15;double *pa[3];for(i=0;i<n;i++)pa=a;if(seidel(pa,b,x,eps,n)) { for(i=0;i<n;i++) printf("\nx[%d] = %g",i,x); }else printf("\nFiled !");}

totoerer · 发表于 2004-5-15 07:43:09

<

center" align=center>梯形法计算定积分 Tixing.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 tixingjf(double a,double b,double eps) { int k=0; double x,h1,h,s1,s2,t1,t2,d; h=0.5*(b-a); s1=(f(a)+f(b))*0.5;s2=f(a+h); t1=h*(s1+s2); while(k<=22) { h1=h;h*=0.5;x=a+h; while(x<b) { s2+=f(x);x+=h1; }t2=h*(s1+s2); if(fabs(t2-t1)<eps)break; k++;t1=t2; } printf("\nk=%d",k); return(t2); }main() { double answer,eps=1.0e-012; answer=tixingjf(0.0,1.0,eps); printf("\nThe Answer is %15.14lf ",answer); }/* 循环18次计算结果为 3.14159265358952 */

totoerer · 发表于 2004-5-15 07:44:14

<

11.8pt; LINE-HEIGHT: 19pt; TEXT-ALIGN: center; mso-char-indent-count: 1.0; mso-char-indent-size: 11.8pt" align=center>辛普生法计算定积分Simpson.c<

19pt">#include<math.h><

19pt">#include<stdio.h>double f(double x) { return(4.0/(1+x*x)); }double simpson(double a,double b,double eps) { int k=0; double x,h1,h2,m1,m2,s2,s1,d,rs; h1=0.5*(b-a); rs=1.0/3.0; m1=f(a)+f(b); m2=f(a+h1); s1=h1*rs*(m1+4.0*m2); while(k<=20){ h2=h1; h1*=0.5; m1+=(m2*2.0); m2=0.0; x=a+h1; while(x<b) { m2+=f(x); x+=h2; } s2=h1*rs*(m1+4.0*m2); if(fabs(s2-s1)<eps)break;s1=s2; k++;} printf("\nk=%d",k); return(s2); }main(){double answer,eps=1.0e-012;answer=simpson(0.0,1.0,eps);printf("\nThe Answer is %15.14lf ",answer);}/* 循环5次计算结果：3.14159265358978 */

totoerer · 发表于 2004-5-15 07:44:37

<

center" align=center>修正牛顿法解非线性方程组newton2.c<

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

21pt; mso-line-height-rule: exactly">#include<stdio.h>#define N 3 double pi,a[N][N],b[N][N],fx[N]; void F(double x[]){double ex,sx;fx[0]=3.0*x[0]-cos(x[1]*x[2])-0.5;fx[1]=x[0]*x[0]-81.0*(x[1]+0.1)*(x[1]+0.1)+sin(x[2])+1.06;ex=exp(-x[0]*x[1]);fx[2]=ex+20.0*x[2]+pi*10.0/3.0-1.0;} void DF(double x[]){double ex,sx;ex=exp(-x[0]*x[1]);a[0][0]=3.0;sx=sin(x[1]*x[2]);a[0][1]=x[2]*sx;a[0][2]=x[1]*sx;a[1][0]=2.0*x[0];a[1][1]=-162.0*(x[1]+0.1);a[1][2]=cos(x[2]);a[2][0]=-x[1]*ex;a[2][1]=-x[0]*ex;a[2][2]=20.0;} int inverta(int n) { int i,j,k,i0; double c,t,maxa,eps=1.0e-15; 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); } int newtonfa(double x0[],double eps){int i,j,k=1;double mx;double x1[N],dx[N],y[N];for(;;) { F(x0); DF(x0); if(!inverta(N))return(0); for(i=0;i<N;i++) { dx=0.0; for(j=0;j<N;j++) dx+=b[j]*fx[j]; } mx=0.0; for(i=0;i<N;i++)if(fabs(dx)>mx)mx=fabs(dx); for(i=0;i<N;i++)y=x0-dx; F(y); for(i=0;i<N;i++) { dx=0.0; for(j=0;j<N;j++) dx+=b[j]*fx[j]; } for(i=0;i<N;i++)x1=y-dx; for(i=0;i<N;i++)x0=x1; if(mx<eps)break; printf("\nk=%d",k); printf("\nx1=%14.12lf x2=%14.12lf x3=%14.12lf",x1[0],x1[1],x1[2]); k++;if(k>400)return(0); }return(1);} main(){int i;double x0[N]={3.0,2.0,1.0},eps=1.0e-12;pi=4.0*atan(1.0);if(newtonfa(x0,eps)) { for(i=0;i<N;i++) printf("\nx[%d] = %14.12lf ",i,x0); }else printf("\nFiled !");}循环5次，计算结果：<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>，<v:shape> <v:imagedata></v:imagedata></v:shape>，<v:shape> <v:imagedata></v:imagedata></v:shape>

totoerer · 发表于 2004-5-15 07:45:31

<

center" align=center>一般迭代法解线性方程组 didai.c<

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

23pt; mso-line-height-rule: exactly">#include<stdio.h> int didai(double *a[],double b[],double x[],double eps,int n){int i,j,k=1;double s,dx,maxdx,x1[80];for(;;) { maxdx=0.0; for(i=0;i<n;i++) { s=b; for(j=0;j<n;j++) s+=(a[j]*x[j]); x1=s; dx=fabs(s-x); if(dx>maxdx)maxdx=dx; } k++; printf("\nk=%d ",k); if(k>200)return(0); if(maxdx<eps)break; for(j=0;j<n;j++)x[j]=x1[j]; }return(1);}main(){int i,n=3;double a[3][3]={0,0.2,0.1,0.2,0,0.1,0.2,0.4,0};double x[3]={1,1,1},b[3]={0.3,1.5,2.0},eps=1.0e-15;double *pa[3];for(i=0;i<n;i++)pa=a;if(didai(pa,b,x,eps,n)) { for(i=0;i<n;i++) printf("\nx[%d] = %g",i,x); }else printf("\nFiled !");}

totoerer · 发表于 2004-5-15 07:46:00

<

>/*已知概率密度，计算概率*/<

> <

>#include<math.h>#include<stdio.h>double fx(double x) { return(exp(-x*x*0.5)); /*<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>*/ }double simpson(double a,double b,double eps,double (*f)(double)) { int m=1; double x,h,h1,sam1,sam2,s2,s1,d,rs; h=0.5*(b-a); rs=1.0/3.0; sam1=f(a)+f(b); sam2=f(a+h); s1=h*rs*(sam1+4.0*sam2); while(m<=19) { h1=h; h*=0.5; sam1+=(sam2*2.0); sam2=0.0; x=a+h; while(x<b) { sam2+=f(x); x+=h1; } s2=h*rs*(sam1+4.0*sam2); if(fabs(s2-s1)<eps)break; s1=s2; m++; } return(s2); }main(){double x=1.0,gailv,eps=1.0e-011,pi,c;pi=4.0*atan(1.0);c=1.0/sqrt(pi*2.0);printf("\nInput x=");scanf("%lf",&x);gailv=c*simpson(-8.0,x,eps,fx); /*<v:shape> <v:imagedata></v:imagedata></v:shape>*/printf("\nF(x)=%15.14lf ",gailv);}

totoerer · 发表于 2004-5-15 07:47:44

<

>上个式子的公式没有弄上去 <

> 我不知道如何在论坛里写公式 <

>请知道的哥们告诉我一 191838238给我论坛短信也行

totoerer · 发表于 2004-5-15 07:49:12

f(x)=e (-(x*x)/2) 是计算公式

		自动登录	找回密码
密码			注-册-帐-号