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文章作者：佚名 责任编辑:admin 发布时间：2012/1/6 15:49:37 |
1、C程序 （1）I显示方法 #include <stdio.h> #define HEIGHT 17 int main(void) { int i=0; printf("\n\nIIIIIII\n"); while (i<HEIGHT) { printf(" III\n"); i++; } printf("IIIIIII\n\n\n"); return 0; } （2）艾特肯 #include<stdio.h> #include<math.h> float f(float a) {float b; b=pow(2,-a); return b; } main() {float x0,x1,e,y,z; int k; scanf("%f,%f",&x0,&e); for(k=0;k<100;k++) {y=f(x0); z=f(y); x1=x0-(y-x0)*(y-x0)/(z-2*y+x0); if(fabs(x1-x0)<e) {printf("%f",x1); break; } else x0=x1; } } （3）杜氏分解法 #include<stdio.h> #include<math.h> main() {int n,i,j,k,t,sum; float a[10][10],b[10],x[10],y[10]; printf("the top exp is "); scanf("%d",&n); for(i=0;i<n;i++) for(j=0;j<n;j++) scanf("%d",&a[i][j]); for(i=0;i<n;i++) scanf("%d",&b[i]); for(i=1;i<n;i++) a[i][1]=a[i][1]/a[1][1]; for(k=1;k<n;k++) {for(j=k;j<n;j++) {sum=0; for(t=0;t<k;t++) sum+=a[k][t]*a[t][j]; a[k][j]=a[k][j]-sum; } if(k!=n-1) {for(i=k+1;i<n;i++) {sum=0; for(t=0;t<k;t++) sum+=a[i][t]*a[t][k]; a[i][k]=(a[i][k]-sum)/a[k][k]; } } } y[0]=b[0]; for(i=1;i<n;i++) {sum=0; for(k=0;k<i;k++) sum+=a[i][k]*y[k]; y[i]=b[i]-sum; } x[n-1]=y[n-1]/a[n-1][n-1]; for(i=n-2;i>=0;i--) {sum=0; for(k=i+1;k<n;k++) sum+=a[i][k]*x[k]; x[i]=(y[i]-sum)/a[i][i]; } for(i=0;i<n;i++) printf("%f\n",x[i]); } （4）二分法 #include<stdio.h> #include<math.h> float f(float x) {float y; y=x*x+2*x+1; return y; } main() {float a,b,c,x; printf("%s","input three numbers:\n"); scanf("%f%f%f",&a,&b,&c); while(fabs(b-a)>=c) {x=(a+b)/2; if(f(x)!=0) {if(f(a)*f(x)<0) b=x; else a=x;} else a=x,b=x; } printf("%f\n",(a+b)/2); } （5）分段线性插值 #include<stdio.h> #include<math.h> main() {int e,h,i,j,k,n; float x[10],y[10],a,sum,total,e1,e2; printf("input n\n"); scanf("%d",&n); printf("input X\n"); for(i=0;i<=n;i++) scanf("%f",&x[i]); printf("input Y\n"); for(i=0;i<=n;i++) scanf("%f",&y[i]); printf("input x\n"); scanf("%f",&a); for(j=1;j<=n;j++) if(x[j-1]<=a&&a<=x[j]) e=j; if(e==1) h=j; else if(e==n) h=j-1; else {e1=fabs(a-x[j-2]); e2=fabs(a-x[j+1]); if(e1<e2) h=j-1; else h=j;} sum=0; for(k=h-1;k<=h+1;k++) {total=1; for(j=h-1;j<k;j++) total*=((a-x[j])/(x[k]-x[j])); for(j=k+1;j<=h+1;j++) total*=((a-x[j])/(x[k]-x[j])); sum+=y[k]*total; } printf("x=%f,L=%f",a,sum); } （6）复合梯形法 #include<stdio.h> #include<math.h> #define f(x) 4/(1+(x)*(x)) float g(int k) {int i,total; total=1; for(i=1;i<=k;i++) total*=2; return total; } main() {int i,k,k0,k1,tag; float a,b,e,e1,t[50],sum; scanf("%f,%f,%f,%d,%d",&a,&b,&e,&k0,&k1); t[0]=((b-a)*(f(1)-f(0)))/2; k=1; tag=0; do{sum=0; for(i=1;i<=g(k-1);i++) sum+=f(a+(2*i-1)*(b-a)/g(k)); t[k]=t[k-1]/2+(b-a)*sum/g(k); if(k>=k1) {e1=fabs(t[k]-t[k-1]); if(e1<e) {printf("k=%d,T=%f\n",k,t[k]); tag=1; } } k++; }while(tag==0&&k<k0); if(tag==0||k>=k0) printf("error!\n"); } （7）复合辛普森 #include<stdio.h> #include<math.h> #define f(x) (x)/(4+(x)*(x)) main() {int k,n; float a,b,h,s,x; scanf("%f,%f,%d",&a,&b,&n); h=(b-a)/(2*n); x=a; s=f(x)-f(b); k=1; do {x=x+h;s=s+4*f(x); x=x+h;s=s+2*f(x); k++; }while(k<=n); s=(s*h)/3; printf("s=%f\n",s); } （8）改进欧拉法 #include<stdio.h> #include<math.h> #define f(x,y) (x)-(y)+1 main() {int k,n; float a,b,h,y0,x,y,t[2]; scanf("%f,%f,%f,%d",&a,&b,&y0,&n); h=(b-a)/n; x=a;y=y0; printf("%f,%f\n",x,y); k=1; do{t[0]=y+h*f(x,y); x=x+h; t[1]=y+h*f(x,t[0]); y=(t[0]+t[1])/2; printf("%f,%f\n",x,y); k++; }while(k<=n); } （9）高斯消去法 #include<stdio.h> #include<math.h> main() {float a[10][10],b[10],m[10][10],x[10],sum; int i,j,k,n; printf("the top exp:"); scanf("%d",&n); printf("\n"); for(i=0;i<n;i++) for(j=0;j<n;j++) scanf("%f",&a[i][j]); for(i=0;i<n;i++) scanf("%f",&b[i]); for(k=0;k<n-1;k++) {if(a[k][k]==0) printf("error"); else for(i=k+1;i<n;i++) {m[i][k]=a[i][k]/a[k][k]; a[i][k]=m[i][k]; b[i]=b[i]-m[i][k]*b[k]; for(j=k+1;j<n;j++) a[i][j]=a[i][j]-m[i][k]*a[k][j]; }} if(a[n-1][n-1]==0) printf("error"); else x[n-1]=b[n-1]/a[n-1][n-1]; b[n-1]=x[n-1]; for(i=n-2;i>=0;i--) {sum=0; for(j=i+1;j<n;j++) {sum+=a[i][j]*x[j];} x[i]=(b[i]-sum)/a[i][i]; b[i]=x[i]; } for(i=0;i<n;i++) printf("%f\n",x[i]); } （10）简单迭代法 #include<stdio.h> #include<math.h> float f(float x) {float y; y=sqrt(10/(4+x)); return y; } main() {float x0,x1,a; int k,N; k=0; scanf("%f,%f,%d",&x0,&a,&N); for(k=0;k<N;k++) {x1=f(x0); if(fabs(x1-x0)<a) {printf("%d,%f",k,x1); break; } else x0=x1; } if(k>=N) {printf("error"); } } （11）列主元元素消元 #include<stdio.h> #include<math.h> main() {float a[10][10],b[10],s,t,e,sum; int i,j,k,n,m; printf("The top exp is "); scanf("%d",&n); for(i=0;i<n;i++) for(j=0;j<n;j++) scanf("%f",&a[i][j]); for(i=0;i<n;i++) scanf("%f",&b[i]); scanf("%f",&e); k=0; do{t=a[k][k]; for(i=k;i<n;i++) {if(fabs(t)<fabs(a[i][k])) {t=a[i][k]; m=i; } else m=k; } if(fabs(t)<e) printf("det A = 0\n"); else {if(m!=k) {for(j=0;j<n;j++) {s=a[m][j]; a[m][j]=a[k][j]; a[k][j]=s; } s=b[m]; b[m]=b[k]; b[k]=s; } for(i=k+1;i<n;i++) for(j=k+1;j<n;j++) {a[i][k]=a[i][k]/a[k][k]; a[i][j]=a[i][j]-a[i][k]*a[k][j]; b[i]=b[i]-a[i][k]*b[k]; } } k++; }while(k<n-2); if(fabs(a[n-1][n-1])<e) printf("det A = 0\n"); else {b[n-1]=b[n-1]/a[n-1][n-1]; for(i=n-2;i>=0;i--) {sum=0; for(k=i+1;k<n;k++) {sum+=a[k][j]*b[j];} b[i]=(b[i]-sum)/a[i][i]; } } for(i=0;i<n;i++) printf("%f\n",b[i]); } （12）龙贝格算法 #include<stdio.h> #include<math.h> #define f(x) 4/(1+(x)*(x)) float g(int k) {int i,total; total=1; for(i=1;i<=k;i++) total*=2; return total; } main() {int k0,k1,k,tag,i; float a,b,e,e1,t[50],s[50],c[50],r[50],sum; scanf("%f,%f,%f,%d,%d",&a,&b,&e,&k0,&k1); t[0]=(b-a)*(f(0)+f(1))/2; k=1; tag=0; do{if(k==1) {t[1]=t[0]/2+(b-a)*f(a+(b-a)/2)/2; s[0]=(4*t[1]-t[0])/3; } if(k==2) {t[2]=t[1]/2+(b-a)*(f(a+(b-a)/4)+f(a+3*(b-a)/4))/4; s[1]=(4*t[2]-t[1])/3; c[0]=(16*s[1]-s[0])/15; } if(k>=3) {sum=0; for(i=1;i<=g(k-1);i++) sum+=f(a+(2*i-1)*(b-a)/g(k)); t[k]=t[k-1]/2+(b-a)*sum/g(k); s[k-1]=(4*t[k]-t[k-1])/3; c[k-2]=(16*s[k-1]-s[k-2])/15; r[k-3]=(64*c[k-2]-c[k-3])/63; } if(k>=k1) {e1=fabs(r[k-3]-r[k-4]); if(e1<e) {printf("k=%d,T=%f\n",k,r[k-3]); tag=1; } } k++; }while(tag==0&&k<k0); if(tag==0||k>=k0) printf("error!\n"); } （13）龙格库塔方法 #include<stdio.h> #include<math.h> #define f(x,y) (x)-(y)+1 main() {int n,k; float a,b,y0,h,x,y,t[4]; scanf("%f,%f,%f,%d",&a,&b,&y0,&n); h=(b-a)/n; x=a;y=y0; printf("%f,%f\n",x,y); k=1; do{t[0]=f(x,y); x=x+h/2; t[1]=f(x,y+h*t[0]/2); t[2]=f(x,y+h*t[1]/2); x=x+h/2; t[3]=f(x,y+h*t[2]); y=y+h*(t[0]+2*t[1]+2*t[2]+t[3])/6; printf("%f,%f\n",x,y); k++; }while(k<=n); } （14）牛顿插值多项式 #include<stdio.h> #include<math.h> main() {int i,j,k,n; float x[10],y[10],a,b,p; printf("input n\n"); scanf("%d",&n); printf("input X\n"); for(i=0;i<=n;i++) scanf("%f",&x[i]); printf("input Y\n"); for(i=0;i<=n;i++) scanf("%f",&y[i]); printf("input x\n"); scanf("%f",&a); b=0; k=0; do{p=1; j=0; do {if(j!=k) p=p*(a-x[j])/(x[k]-x[j]); j++; }while(j<n); b=b+p*y[k]; k++; }while(k<n); printf("x=%f,y=%f",a,b); } （15）牛顿迭代 #include<stdio.h> #include<math.h> #define N 100 float f(float x) {float y; y=x-cos(x); return y; } main() {float x0,x1,a; int k; scanf("%f,%f",&x0,&a); for(k=0;k<N;k++) {x1=x0-(x0-cos(x0))/(1+sin(x0)); if(fabs(x1-x0)<a) {printf("%f,%d",x1,k); break; } else x0=x1; } if(k>=N) {printf("error"); } } （16）牛顿下山 #include<stdio.h> #include<math.h> float f(float x) {float y; y=pow(x,3)-x-1; return y; } float g(float x) {float y; y=3*pow(x,2)-1; return y; } main() {float x0,x1,b,e,a,x; int k; scanf("%f,%f",&x0,&a); k=0; if(fabs(g(x0))<a) {printf("error"); }else e=1.00; while(fabs(f(x0))>=a) {do{x1=x0-e*f(x0)/g(x0); b=x0; x0=x1; e=e/2.00; }while(fabs(f(x1))>=fabs(f(b))); k+=1;e=1.00; } x=x0; printf("%d,%f",k,x); } （17）秦九韶 #include<stdio.h> #include<math.h> #define N 100 main() {int m,i,j; double sum,x,a[N]; printf("input the x and the top exp:"); scanf("%lf,%d",&x,&m); printf("\ninput the a[i](0<i<100):"); for(i=0;i<=m;i++) {scanf("%lf",&t); a[i]=t; } sum=a[m]; for(j=m-1;j>=0;j--) sum=x*sum+a[j]; printf("\nthe result is:%lf\n",sum); } （18）三对角线追赶法 #include<stdio.h> #include<math.h> main() {int i,j,k,n; float d[10][10],g[10],a[10],b[10],c[10],x[10],y[10],f[10]; printf("the top exp is "); scanf("%d",&n); scanf("%f,%f,%f,%f",&d[0][0],&d[0][1],&d[n-1][n-2],&d[n-1][n-1]); for(i=1;i<n-1;i++) for(j=i-1;j<=i+1;j++) scanf("%f",&d[i][j]); for(i=0;i<n;i++) scanf("%f",&g[i]); for(i=1;i<n-1;i++) a[i]=d[i][i-1]; for(i=0;i<n;i++) b[i]=d[i][i]; for(i=0;i<n-1;i++) c[i]=d[i][i+1]; f[0]=c[0]/b[0]; for(k=1;k<n-1;k++) f[k]=c[k]/(b[k]-a[k]*f[k-1]); y[0]=g[0]/b[0]; for(i=1;i<n;i++) y[i]=(g[i]-a[i]*y[i-1])/(b[i]-a[i]*f[i-1]); x[n-1]=y[n-1]; for(i=n-2;i>=0;i--) x[i]=y[i]-f[i]*x[i+1]; for(i=0;i<n;i++) printf("%f\n",x[i]); } （19）系统 #include <stdio.h> int main(void) { int a,b,CHECK; printf("Input two nonzero intergers:"); scanf("%d,%d",&a,&b); printf("%s%4d\n%s%4d\n%s%4d\n%s%4d\n%s%4d\n", " a=",a, " b=",b, " a/b=",a/b, " a%b=",a%b, " CHECK=",(a/b)*b+a%b-a ); return 0; } （2）CHAIN：马尔可夫链算法 #include <stdio.h> #include <math.h> double a[10][10]; void Guass(int n){ int i,j,k; double t; for(k=0;k<n-1;k++) {t=a[k][k]; for(j=k;j<n;j++) a[k][j]=a[k][j]/t; for(i=0;i<n-1;i++) if(i!=k) {t=a[i][k]/a[k][k]; for(j=k;j<n;j++) a[i][j]=a[i][j]-a[k][j]*t; } } return; } void chain(){ static double p[10][10],pr[10],diff,table[100][10],pnew[10][10],ptemp[10][10],temp[10],exr[10][10]; int n,i,j,k,s,m,found,inr,inc; printf("Welcome to the MARKOV CHAIN ANALYSIS system!\n\n"); printf("how many states =? "); scanf("%d",&n); printf("\nthe steady transmit possibility of step 1 ?\n"); for(i=0;i<n;i++) for(j=0;j<n;j++) scanf("%lf",&p[i][j]); printf("\nthe initiate state of step 1 ?\n"); for(i=0;i<n;i++) scanf("%lf",&pr[i]); for(i=0;i<n;i++) for(j=0;j<n;j++) pnew[i][j]=p[i][j]; for(i=0;i<n;i++) table[0][i]=pr[i]; printf("\n step 1"); for(i=0;i<n;i++) {printf("\n"); for(j=0;j<n;j++) printf("%f\t",p[i][j]); } printf("\n"); for(k=2;k<100;k++) {for(j=0;j<n;j++) {temp[j]=0; for(i=0;i<n;i++) temp[j]=temp[j]+pr[i]*pnew[i][j]; } for(i=0;i<n;i++) table[k-1][i]=temp[i]; for(i=0;i<n;i++) for(j=0;j<n;j++) {ptemp[i][j]=0; for(m=0;m<n;m++) ptemp[i][j]=ptemp[i][j]+p[i][m]*pnew[m][j]; } for(i=0;i<n;i++) for(j=0;j<n;j++) pnew[i][j]=ptemp[i][j]; for(j=0;j<n;j++) {for(i=0;i<n-1;i++) { for(m=i+1;m<n;m++) {diff=pnew[i][j]-pnew[m][j]; if(diff<0) diff=-diff; if(diff>0.001) {found=0; break; } found=1; } if(diff>0.001) break; } if(diff>0.0001) break; } if(found==0) {if(k%5==0) {printf("\n step %d",k); for(i=0;i<n;i++) {printf("\n"); for(j=0;j<n;j++) printf("%f\t",pnew[i][j]); } getch(); } if(k>=100) {printf("\nsteady_state probability have not been detained in 100"); return; } } else {printf("\nstep %d",k); for(i=0;i<n;i++) {printf("\n"); for(j=0;j<n;j++) printf("%f\t",pnew[i][j]); } break; } } for(j=0;j<n;j++) {temp[j]=0; for(i=0;i<n;i++) temp[j]=temp[j]+pr[i]*pnew[i][j]; } for(i=0;i<n;i++) table[k][i]=temp[i]; printf("\nThe steady-state probability of being in \n"); for(j=0;j<n;j++) printf("state %d is %f\n",j,pnew[n-1][j]); printf("probability of being in state\n\n"); for(i=0;i<=k;i++) {printf("%d",i); for(j=0;j<n;j++) printf("\t%f",table[i][j]); printf("\n"); if(i%10==0) getch(); } for(s=0;s<n;s++) { inr=0; for(j=0;j<n;j++) if(j==s) continue; else {inc=0; for(i=0;i<n;i++) if(i==s) continue; else { a[inr][inc]=-p[j][i]; if(j==i) a[inr][inc]=1+a[inr][inc]; inc++; } inr++; } for(i=0;i<n-1;i++) a[i][n-1]=1; Guass(n); i=0; for(j=0;j<n;j++) if(j!=s) exr[j][s]=a[i++][n-1]; else exr[j][s]=1/pnew[n-1][s]; } printf("\nTable of expected first passage times and recurrence times"); for(i=0;i<n;i++) {printf("\n%d",i); for(j=0;j<n;j++) printf("\t%f",exr[i][j]); } } void main(){ clrscr(); chain(); getch(); } （3）DECISION：贝叶斯决策方法 #include <stdio.h> #include <math.h> #define pi 3.14159 #define p(x,t) exp(-(x-t)*(x-t)/20)/sqrt(20*pi) void decision(){ int i,j,type,m,n,flag,state[5],index; float xx,a[5][5],p[5],e[5],sum,decision; printf("Welcome to the DECISION_STSTEM!"); printf("\n\ntype of the problem,max(key '0')or min(key '1')? "); scanf("%d",&type); printf("type of the decision,without data(key'0')or with data(key'1')? "); scanf("%d",&flag); printf("number of the actions "); scanf("%d",&m); printf("number of the nature states "); scanf("%d",&n); for(j=0;j<n;j++) printf("\tt%d",j+1); printf("\n"); for(i=0;i<m;i++) {printf("A%d\t",i+1); for(j=0;j<n;j++) scanf("%f",&a[i][j]); } printf("probability of the natural states=?\n"); for(j=0;j<n;j++) scanf("%f",&p[j]); if(flag) {printf("given data=? "); scanf("%f",&xx); printf("states of nature=?\n"); for(j=0;j<n;j++) scanf("%d",&state[j]); sum=0; for(j=0;j<n;j++) sum+=p[j]*p(xx,state[j]); for(j=0;j<n;j++) p[j]=p[j]*p(xx,state[j])/sum; } for(i=0;i<m;i++) {sum=0; for(j=0;j<n;j++) sum+=p[j]*a[i][j]; e[i]=sum; } decision=e[0]; index=1; if(type==0) for(i=1;i<m;i++) { if(decision<e[i]) {decision=e[i]; index=i+1; } } else for(i=0;i<m;i++) if(decision>e[i]) {decision=e[i]; index=i+1; } printf("\n**********\n"); printf("Results:"); printf("\n**********\n"); printf("expected loss for each course of action based on prior distribution\n"); for(i=0;i<m;i++) printf("%f\t",e[i]); printf("\n\nThe optimum expected loss is %f ",decision); printf("\n\nchoose action A( %d )",index); return; } void main(){ clrscr(); decision(); getch(); } （4）dp_invest：动态规划的投资问题 #include <stdio.h> int istar[10]; float dp_invest(int N,int K){ int i,j,sum,z,d[10][50]; float g[10][50],f[10][50]; printf("\nThe return function values as follows!\n"); for(j=0;j<K+1;j++) printf("\t%d",j); for(i=0;i<N;i++) {printf("\ntask%d:\t",i+1); for(j=0;j<K+1;j++) scanf("%f",&g[i][j]); } for(i=0;i<N;i++) for(j=0;j<K+1;j++) {f[i][j]=0;d[i][j]=0;} for(j=0;j<K+1;j++) {f[N-1][j]=g[N-1][j]; d[N-1][j]=j; } for(i=N-2;i>=0;i--) for(j=1;j<K+1;j++) {f[i][j]=g[i][0]+f[i+1][j]; d[i][j]=0; for(z=1;z<=j;z++) if((g[i][z]+f[i+1][j-z])>f[i][j]) {f[i][j]=g[i][z]+f[i+1][j-z]; d[i][j]=z; } } istar[0]=d[0][K]; for(i=1;i<N;i++) {sum=0; for(j=0;j<=i-1;j++) sum=sum+istar[j]; istar[i]=d[i][K-sum]; } return f[0][K]; } void main(){ int i,N,K; clrscr(); printf("WELCOME TO THE DYNAMIC_INVEST SYSTEM!\n\n"); printf("How many tasks ? "); scanf("%d",&N); printf("\nHow many units of materials ? "); scanf("%d",&K); printf("\nThe optimal return is %f",dp_invest(N,K)); for(i=0;i<N;i++) printf("\nThe optimal amout to invest in task %d is %d",i+1,istar[i]); getch(); } （5）dp_plan：生产计划算法 #include <stdio.h> #define pc(j) 20+5*j #define e(j) j void dp_plan(){ int i,j,k,sum,limit,n,io,max_stock,max_production,d[10],x[10][100],z,z_maxlimit,z_minlimit,xstar[10]; float f[10][100],temp; printf("WELCOME TO THE DYNIMIC SYSTEM!\n\nperiods of the inventory =? "); scanf("%d",&n); printf("\nthe maximum stocks of each period=? "); scanf("%d",&max_stock); printf("\nthe maximum production of each period=? "); scanf("%d",&max_production); printf("\nThe stocks of the first period=? "); scanf("%d",&io); for(i=0;i<n;i++) {printf("Demand for period %d is ",i+1); scanf("%d",&d[i]); } for(k=0;k<d[n-1];k++) {x[n-1][k]=d[n-1]-k; f[n-1][k]=pc(x[n-1][k]); } f[n-1][d[n-1]]=0; x[n-1][d[n-1]]=0; for(i=n-2;i>=0;i--) {sum=0; for(j=i;j<n;j++) sum+=d[j]; if(sum<max_stock) limit=sum; else limit=max_stock; for(k=0;k<=limit;k++) {if(d[i]-k>0) {z_minlimit=d[i]-k; f[i][k]=pc(z_minlimit)+e(0)+f[i+1][0]; x[i][k]=z_minlimit; } else {z_minlimit=0; f[i][k]=e(k-d[i])+f[i+1][k-d[i]]; x[i][k]=0; } if(sum-k>max_stock+d[i]-k) if(max_stock+d[i]-k>max_production) z_maxlimit=max_production; else z_maxlimit=max_stock+d[i]-k; else if(sum-k>max_production) z_maxlimit=max_production; else z_maxlimit=sum-k; for(z=z_minlimit;z<=z_maxlimit;z++) {temp=pc(z)+e(k+z-d[i])+f[i+1][k+z-d[i]]; if(f[i][k]>temp) {f[i][k]=temp; x[i][k]=z; } } } } /* for(i=0;i<n;i++) {printf("\n\nthe period %d\n",i+1); for(j=0;j<=5;j++) printf("\t%f--->%d\t",f[i][j],x[i][j]); getch(); }*/ printf("\n\nThe minimum policy cost for the %d periods is %f",n,f[0][io]); xstar[0]=x[0][io]; j=io; printf("\n\nThe optimal amount to produce in period 1 is %d",xstar[0]); for(i=1;i<n;i++) {xstar[i]=x[i][xstar[i-1]-d[i-1]+j]; printf("\nThe optimal amount to produce in period %d is %d",i+1,xstar[i]); j=xstar[i-1]-d[i-1]+j; } } void main(){ clrscr(); dp_plan(); getch(); } （6）linear：线性规划单纯形算法 #include <stdio.h> int K,M,N,Q=100,Type,Get,Let,Et,Code[50],XB[50],IA,IAA[50],Indexg,Indexl,Indexe; float Sum,A[50][50],B[50],C[50]; void initiate(); void solve(); void main(){ int i,j; clrscr(); /****** input data ******/ printf("\nWelcome to the linear programming solution!\n\n********\nNotice 1!\n********\nIf the type of objective equation is 'max' ,please enter '1'!\nElse please enter '0'!\n\n\n"); printf("********\nNotice 2!\n********\nDefine the type of subjective equation as following:\n'<='is equal to '0'!\n'>=' is equal to '1'!\n'='is equal to '2'!\n"); printf("\n\nPlease input the coefficients or the constants:\n"); printf("THe number of subjective equations ? "); scanf("%d",&M); printf("THe number of variables ? "); scanf("%d",&K); printf("THe number of ' <= 'subjective equations ? "); scanf("%d",&Let); printf("THe number of ' >= 'subjective equations ? "); scanf("%d",&Get); printf("THe number of ' = 'subjective equations ? "); scanf("%d",&Et); printf("THe type of objective equation ? "); scanf("%d",&Type); N=K+Let+Et+2*Get; for(i=0;i<M;i++) {printf("Please input %d's equation:\n",i+1); printf("type ? "); scanf("%d",&Code[i]); printf("constant ? "); scanf("%f",&B[i]); for(j=0;j<K;j++) {printf("coefficient ? "); scanf("%f",&A[i][j]); } } printf("Plese input the constants of the object equation:\n"); for(j=0;j<K;j++) scanf("%f",&C[j]); printf("\n\n*************************************\nPlease check the data you just input!\n************************************* \n"); getch(); if(Type) printf("The type of the object equation is 'max'\n"); else printf("The type of the object equation is 'min'\n"); printf("\nThe object equation is :\n"); for(j=0;j<K;j++) {printf("(%f)X%d ",C[j],j+1); if(j!=K-1) printf("+"); } printf("\n\nThe suject equation is :"); for(i=0;i<M;i++) {printf("\nNumber %d suject equation is: %d %f ",i+1,Code[i],B[i]); for(j=0;j<K;j++) {printf("(%f)X%d ",A[i][j],j+1); if(j!=K-1) printf("+"); } } /****** initiate data ******/ initiate(); solve(); if(!Type) A[M][N]=-A[M][N]; printf("\n\nThe optimal value of the original objective function is: %f ",A[M][N]); getch(); } /****** initiate variables function ******/ void initiate(){ int i,j; Indexg=K; Indexl=Indexg+Get; Indexe=Indexl+Let;; for(i=0;i<M+1;i++) for(j=K;j<N+1;j++) A[i][j]=0; for(i=0;i<M;i++) A[i][N]=B[i]; for(i=0;i<M;i++) switch(Code[i]) { case 0: {XB[i]=Indexl; A[i][Indexl++]=1; break; }; case 1: {XB[i]=Indexe; IAA[IA++]=i; A[i][Indexe++]=1; A[i][Indexg++]=-1; break; }; case 2: {XB[i]=Indexe; IAA[IA++]=i; A[i][Indexe++]=1; break; }; } for(j=0;j<K;j++) if(Type) A[M][j]=-C[j]; else A[M][j]=C[j]; for(j=K;j<=N;j++) A[M][j]=0; for(j=K+Get+Let;j<N;j++) A[M][j]=Q; Sum=0; for(j=0;j<=N;j++) {Sum=0; for(i=0;i<IA;i++) Sum=Sum+A[IAA[i]][j]; A[M][j]=A[M][j]-Sum*Q; } return; } /****** process data function ******/ void solve(){ int i,j,mark=1,minus,minusmark,basic=0,divide,dividemark; float H,P; while(1) {mark=0; minusmark=0; minus=0;divide=1000; dividemark=0; printf("\n\nBasic solution %d is\n",++basic); for(i=0;i<M;i++) printf("Basic variable %d = X( %d )= %f\n",i+1,XB[i]+1,A[i][N]); printf("\nCurrent value of the object equation is: %f\n",A[M][N]); getch(); for(j=0;j<N;j++) { if(A[M][j]<-6e-8) mark++; if(A[M][j]<minus) {minus=A[M][j]; minusmark=j; } } if(mark==0) break; for(i=0;i<M;i++) {if(A[i][minusmark]==0) continue; if(A[i][N]/A[i][minusmark]<=0) continue; if(A[i][N]/A[i][minusmark]<divide) {divide=A[i][N]/A[i][minusmark]; dividemark=i; } } XB[dividemark]=minusmark; if(divide<0) printf("There is no solution because of no boundary!"); P=A[dividemark][minusmark]; for(j=0;j<N+1;j++) A[dividemark][j]=A[dividemark][j]/P; for(i=0;i<M+1;i++) {H=A[i][minusmark]; if(i==dividemark) continue; for(j=0;j<N+1;j++) A[i][j]=A[i][j]-H*A[dividemark][j]; } } printf("\n\n*************************************\nThe last basic solution is optimal!\n************************************* \n"); return; } |
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