根据离散分布生成随机数——通用例子

//////////////////////////////////////////////////////how to create discrete random number according //to you giving distribution////auther :Hui Li 2016/3/16 00:22 ////////////////////////////////////////////////////int f02(){ //double a[5]={1.5,2.5,3.5,4.5,5.5};double
p[5]={0.1,0.3,0.1,0.4,0.1};double x,content,C;double N=10000;//change log:change the range ,this program should be modifyeddouble a=0,b=500,c=5;//TH1D* h1=new TH1D("h1","h1;x;y",c,a,b);TH1D* h2=new TH1D("h2","h2;x;y",c,a,b); double sum(double p[],int j);TRandom3
r;for(int i=0;i<N;i++){x=r.Uniform();for(int j=0;j<5;j++){ if(sum(p,j)<x&&x<sum(p,j+1)){ x=(x+j)*(b-a)/c;//cout<<"\tj="<<j<Fill(x);}}}for (Int_t i=1;i<=5;i++) { count=h1->GetBinContent(i); C=count/N; h2->SetBinContent(i,C);} // //归一化 normalization// for(int
j=1;j<=5;j++) // { // content =h1 ->GetBinContent(j); // C=content/N; // h2 ->SetBinContent(j,C);// } auto c1=new TCanvas(); c1->Divide(1,2); c1->cd(1); h1->Draw(); c1->cd(2); h2->Draw(); return 0;} double sum(double a[],int j){ double sum=0; for(int i=0;i<j;i++)
sum+=a[i]; return sum;}