C语言解题经典程序

totoerer · 发表于 2004-5-14 08:12:11

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align=center>C语言常用数学类库函数
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>1. <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 connecttype="rect" gradientshapeok="t" extrusionok="f"></v:path><lock aspectratio="t" v:ext="edit"></lock></v:shapetype><v:shape><v:imagedata></v:imagedata></v:shape> 求整型数x的绝对值
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>2. double fabs(double x); 求浮点数x的绝对值
3. double sqrt(double x); 求x的平方根
4. double exp(double x); 求<v:shape> <v:imagedata></v:imagedata></v:shape>的值
5. double log(double x); 求ln x的值
6. double log10(double x); 求log 10 x的值
7. double pow(double x, double y); 求x的y次方的值
8. 三角函数
 double sin(double x); 求sin x的值
 double cos(double x); 求cos x的值
 double tan(double x); 求tg x的值
9. 反三角函数
 double asin(double x); 求arcsin x的值
 double acos(double x); 求arccos x的值
 double atan(double x); 求arctg x的值
10. int rand(void); 产生一个伪随机数
11. void srand(unsigned sd); 用参数sd初始化随机数发生器

totoerer · 发表于 2004-5-14 08:12:44

<

center" align=center>阿达姆斯预报-校正法解微分方程 ADAMS.C<

18pt; mso-line-height-rule: exactly">#include<math.h><

18pt; mso-line-height-rule: exactly">#include<stdio.h> double f(double x,double y){ return(x*exp(-y)); } double rungkuta(double x0,double y0,double h) { double h12,h16,x,y,K1,K2,K3,K4; h12=0.5*h; h16=h/6.0; x=x0; y=y0; K1=f(x,y); x+=h12; y=y0+h12*K1; K2=f(x,y); y=y0+h12*K2; K3=f(x,y); x+=h12; y=y0+h*K3; K4=f(x,y); y=y0+(K1+2.0*K2+2.0*K3+K4)*h16; return(y); } double get_step(double x0,double y0,double xn,double eps) { int k=0; double h,y,y1,y2; h=(xn-x0)/100.0; y=rungkuta(x0,y0,h); while(k<32) { h*=0.5; y1=rungkuta(x0,y0,h); y2=rungkuta(x0+h,y1,h); if(fabs(y2-y)<eps)break; y=y1;k++; } return(h); } double adams(double x0,double y0,double xn,double eps) { int i; unsigned int M; double h,h24,x,y; double yk0,yk1,yk2,yk3,yk4,yk40; double fk0,fk1,fk2,fk3,fk4; h=get_step(x0,y0,xn,eps); x=x0; yk1=rungkuta(x,y0,h);x+=h; yk2=rungkuta(x,yk1,h);x+=h; yk3=rungkuta(x,yk2,h); h24=h/24.0;x=x0;yk0=y0; fk0=f(x,yk0); x+=h;fk1=f(x,yk1); x+=h;fk2=f(x,yk2); x+=h;fk3=f(x,yk3); for(;;) { yk4=yk3+h24*(55.0*fk3-59.0*fk2+37.0*fk1-9.0*fk0); x+=h;fk4=f(x,yk4); yk4=yk3+h24*(9.0*fk4+19.0*fk3-5.0*fk2+fk1); if(fabs(x-xn)<=h*0.2)break; fk0=fk1;fk1=fk2;fk2=fk3; fk3=f(x,yk4); yk3=yk4; } return(yk4); } main() { double answer,eps=1.0e-16; answer=adams(0.0,0.0,5.0,eps); printf("\nThe Answer is %16.15lf",answer); }

totoerer · 发表于 2004-5-14 08:13:16

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center" align=center>变步长搜索法求函数的极大值fv.c<

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

22pt; mso-line-height-rule: exactly">#include<stdio.h>#include<conio.h>double f(double x){return(x*(2.0-log(x)));}double fv(double a,double b,double eps){double h,x0,x1,f0,f1;int n=0;h=(b-a)*0.25;x0=a;f0=f(x0);for(;;) { x1=x0+h;if(x1>b)x1=b;if(x1<a)x1=a; f1=f(x1); if(f1>f0) { x0=x1;f0=f1;h*=2.0; } else if(f1==f0) { x0=(x0+x1)*0.5; f0=f(x0); h*=(-0.25); } else { h*=(-0.25); } if(fabs(h)<eps)break; n++; }printf("\n n=%d",n);return(x0);} main(){double mx,my;mx=fv(1.0,4.0,1.0e-10);my=f(mx);printf("\n Maxx=%14.12lf Maxy=%14.12lf",mx,my);getch();} 计算结果：n=47Maxx=2.718281804060Maxy=2.718281828459

totoerer · 发表于 2004-5-14 08:20:16

<

center" align=center>不动点迭代法解非线性方程组DIEDAIF.C<

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

24pt; mso-line-height-rule: exactly">#include<stdio.h>double g1(double x[]){return(sqrt(0.5*(x[0]*(x[1]+5)-1)));}double g2(double x[]){return(sqrt(x[0]+3*log10(x[0])));} int diedai(double x0[],double eps,int n){int i,k=1;double dx,maxdx,x1[80]; for(;;) { x1[0]=g1(x0);x1[1]=g2(x0); maxdx=0.0; printf("\nk=%d ",k); for(i=0;i<n;i++) printf("x[%d]=%12.10lf ",i,x1); getch(); for(i=0;i<n;i++) { dx=fabs(x1-x0); if(dx>maxdx)maxdx=dx; } k++;if(k>200)return(0); if(maxdx<eps)break; for(i=0;i<n;i++)x0=x1; }return(1);} main(){int i,n=2;double x0[2]={3,2};if(diedai(x0,1.0e-12,n)) { for(i=0;i<n;i++) printf("\nx[%d] = %12.10lf",i,x0); }else printf("\nFiled !");}

totoerer · 发表于 2004-5-14 08:23:01

<

center" align=center>从图中找出哈密顿回路<

>#include<stdio.h><

>#define NE 7void getline(int ne,int *pd2[],int *de,int *me,int pp,int pe,int *no){int i,j,h;if(pe==ne&&de[ne-1]==0) { ++(*no); printf("\nNO%5d methed: ",*no);printf(" 0 "); for(i=0;i<ne;++i)printf(" %d ",de); printf("\n"); return; }for(j=0;j<ne;++j) { if(pd2[pp][j]!=0&&me[j]==0) { me[j]=1;de[pe]=j; getline(ne,pd2,de,me,j,pe+1,no); printf(" no=%d ",*no); me[j]=0; } }} main(){int ne,*pd2[NE],i,j,h,pp,pe,de[NE],me[NE],no;int a[NE][NE]={{0,1,1,0,0,0,0},{1,0,1,1,0,0,0},{1,1,0,0,1,1,0},{0,1,0,0,0,1,0},{0,0,1,0,0,0,1},{0,0,1,1,0,0,1},{0,0,0,0,1,1,0}};ne=NE;for(i=0;i<ne;i++)pd2=&a[0];for(i=0;i<ne;i++)me=0;pp=0;pe=0;no=0;getline(ne,pd2,de,me,pp,pe,&no);printf("\n* * * total methed is %d * * *",no);}

totoerer · 发表于 2004-5-14 08:23:31

<

21pt; TEXT-ALIGN: center; mso-line-height-rule: exactly" align=center>/*-----大M法求解线性规划-----*/<

>#include<stdio.h><

>#include<math.h>main() { int i,j,k,r,s,m=3,n=2,h=7,num=0, jbl[3]={2,5,6}; double a[3][8]={1,1.2,1,0,0,0,0,12,6,1,0,-1,0,1,0,12,1,2,0,0,-1,0,1,16}; double c[8], zeta[3]={0,0,0}, M=1.0e10,maxc,zr,d,f=0.0; c[0]=4+7*M; c[1]=3+3*M; c[2]=0; c[3]=-M; c[4]=-M;c[5]=0;c[6]=0;c[7]=28.0*M; while(num<=h) { maxc=c[0];s=0; for(j=1;j<h;j++) if(c[j]>maxc) { maxc=c[j];s=j; } if(maxc<=0.0)break; for(i=0;i<m;i++) if(a>0.0)zeta=a[h]/a; else zeta=M; zr=zeta[0];r=0; for(i=1;i<m;i++) if(zeta<zr) { zr=zeta;r=i; } jbl[r]=s; d=a[r];for(j=0;j<=h;j++) a[r][j]/=d; a[r]=1.0; for(i=0;i<m;i++) if(i!=r) { d=a; for(j=0;j<=h;j++) a[j]-=a[r][j]*d; a=0.0; } d=c; for(j=0;j<=h;j++) c[j]-=a[r][j]*d; c=0.0; num++; } for(i=0;i<m;i++) printf("\nx%d=%g",jbl+1,a[h]); printf("\nMax f=%g",fabs(c[h])); }

totoerer · 发表于 2004-5-14 08:23:59

<

center" align=center>/*-----单纯形法解线性规划-----*/<

14pt; mso-line-height-rule: exactly">#include<stdio.h><

14pt; mso-line-height-rule: exactly">#include<math.h>main() { int i,j,k,r,s,m=3,n=2,h=5,num=0, jbl[3]={0,0,0}; double a[3][6]={2,9,1,0,0,18,2,4,0,1,0,10,3,2,0,0,1,12}; double b[3]={18,10,12}, c[6]={3,4,0,0,0,0}; double zeta[3]={2,3,4}, M=5.0e10,maxc,zr,d; while(num<=h) { maxc=c[0];s=0; for(j=1;j<h;j++) if(c[j]>maxc) { maxc=c[j];s=j; } if(maxc<=0)break; for(i=0;i<m;i++) if(a>0.0)zeta=a[h]/a; else zeta=M; zr=zeta[0];r=0; for(i=1;i<m;i++) if(zeta<zr) { zr=zeta;r=i; } if(zr==M){ printf(“No solution !”);exit(1); } jbl[r]=s;d=a[r]; for(j=0;j<=h;j++) a[r][j]/=d; a[r]=1.0; for(i=0;i<m;i++) if(i!=r) { d=a; for(j=0;j<=h;j++) a[j]-=a[r][j]*d; a=0.0; } d=c; for(j=0;j<=h;j++) c[j]-=a[r][j]*d; c=0.0; num++; } for(i=0;i<m;i++) printf("\nx%d=%g",jbl+1,a[h]); printf("\nMax f=%g",fabs(c[h])); }

totoerer · 发表于 2004-5-14 08:24:38

<H2 center" align=center>等距离法求函数的极大值dengq.c</H2><

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

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

18pt">double f(double x){return(x*sin(x));}double ca_max(double a,double b,double eps){unsigned int n=0;double c,x1,x2,x3,f1,f2,f3;c=(b-a)*0.25;x1=a+c;f1=f(x1);x2=x1+c;f2=f(x2);x3=x2+c;f3=f(x3);for(;;) { if(f1>=f2&&f1>f3) { b=x2;x2=x1;f2=f1; } else if(f3>=f2&&f3>f1) { a=x2;x2=x3;f2=f3; } else { a=x1;b=x3; } c=(b-a)*0.25; if(fabs(c)<eps)break; x1=a+c;f1=f(x1); x3=x2+c;f3=f(x3); n++; }printf("\nn=%u",n);return(x2);} main(){double mx,my;mx=ca_max(0.0,3.16,1.0e-12);my=f(mx);printf("\n Maxx=%14.12lf Maxy=%14.12lf",mx,my);} 计算结果：n=39Maxx=2.028757837284Maxy=1.819705741160

totoerer · 发表于 2004-5-14 08:25:08

<

center" align=center>等距离法求函数的极小值<

>#include<math.h><

>#include<stdio.h>#include<conio.h>double fx(double x){return(sin(x)-log(1+x*x));}double ca_min(double a,double b,double eps){double c,x1,x2,x3,f1,f2,f3;c=(b-a)*0.25; x1=a+c; f1=fx(x1);x2=x1+c; f2=fx(x2); x3=x2+c; f3=fx(x3);for(;;) { if(f1<=f2&&f1<f3) { b=x2; x2=x1; f2=f1; } else if(f3<=f2&&f3<f1) { a=x2; x2=x3; f2=f3; } else { a=x1;b=x3; } c=(b-a)*0.25; if(fabs(c)<eps)break; x1=a+c;f1=fx(x1); x3=x2+c;f3=fx(x3); }return(x2);}main(){double mx,my;mx=ca_min(4.0,6.0,1.0e-10); my=fx(mx);printf("\n Maxx=%14.12lf Maxy=%14.12lf",mx,my); getch();}

tangjianye · 发表于 2004-5-20 01:26:26

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>好东东！！！！！！！！

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C语言解题经典程序

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