UVA&&POJ discrete probability exercise [3]

POJ3869 headshot

Test instructions: Give the revolver bullet sequence, hit a gun no bullets, to make the next shot is not the maximum probability should rotate or shoot

Conditional probability, |00|/(|00|+|01|) And |0|/n who the big question

|00|+|01|=|0|

Note that the sequence is a ring

////main.cpp//poj3869////Created by Candy on 25/10/2016.//copyright©2016 Candy. All rights reserved.//#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespacestd;Const intn= the;intn,a,b;CharS[n];intMainintargcConst Char*argv[]) {scanf ("%s", s+1); N=strlen (s+1);  for(intI=1; i<=n;i++){        if(s[i]=='0') b++; if(s[i]=='0'&&s[i+1]=='0') a++; }    if(s[n]=='0'&&s[1]=='0') a++; if(a*n>b*b) printf ("SHOOT"); Else if(a*n==b*b) printf ("EQUAL"); Elseprintf"ROTATE"); return 0;}

UVA-10491

Cows and Cars

Classic problem, a cow, b car, C door Show

Full probability formula, divided into the first selected cattle A/(A+B) and the first selected car B/(a+b) two parts

Part of the always change the door to get the car probabilities are b/(a+b-c-1) and (b-1)/...

 #include <cstdio>double   A,b,c;  int  Main (int  argc, const  char  * argv[]) { while  (scanf ( %lf%lf%lf   ", &a,&b,&c)!=eof) printf (  " %.5f\n  , (a*b+b* (B-1 ))/((a+b) * (A+b-c-1   return  0  ;}  

uva-11181probability| Given

Test instructions: N person, buy things probability pi, have r personal buy something, ask everyone actually buy things probability

Conditional probabilities

E for R personal buy things, ei for R personally I buy something

P (ei| e) =p (EiE)/P (E)

Calculate probability with Dfs pop-up everyone buy or not buy

PS: Do not read into optimization, floating point number

////main.cpp//uva11181////Created by Candy on 25/10/2016.//copyright©2016 Candy. All rights reserved.//#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespacestd;Const intn= -; typedefLong LongLl;inlineintRead () {CharC=getchar ();intx=0, f=1;  while(c<'0'|| C>'9'){if(c=='-') f=-1; c=GetChar ();}  while(c>='0'&&c<='9') {x=x*Ten+c-'0'; c=GetChar ();} returnx*F;}intN,r,vis[n];DoubleP[n],pe[n];voidDfsintDintCntDoubleProb) {//printf ("Dfs%d%d%f\n", d,cnt,prob);    if(cnt>r| | D-1-CNT&GT;N-R)return; if(d==n+1){        if(cnt==R) for(intI=1; i<=n;i++)if(Vis[i]) pe[i]+=prob; pe[0]+=prob; return; } Vis[d]=1; DFS (d+1, cnt+1, prob*P[d]); VIS[D]=0; DFS (d+1, cnt,prob* (1-p[d]));}intMainintargcConst Char*argv[]) {    intcas=0;  while((N=read ())) {printf ("Case %d:\n",++CAs); R=read ();  for(intI=1; i<=n;i++) scanf ("%LF",&P[i]); memset (Vis,0,sizeof(VIS)); memset (PE,0,sizeof(PE)); DFS (1,0,1.0);  for(intI=1; i<=n;i++) printf ("%.6f\n", pe[i]/pe[0]); }        return 0;}

UVA&&POJ discrete probability exercise [3]