"ZOJ" "3329" one person Game

Probability DP

Kuangbin summed up the third way in the title

It seems that there is no state of entry ah ... It's all about DP. Of course, to find "the transition between states" ...

In the subject Dp[i] and dp[i-(K1+K2+K3)] to Dp[i-1] all have relations ... Then all the dp[i] are related to dp[0] that is ans ...

Use "factor" to transfer ... sigh has been getting more and more water lately.

1 //Bzoj2#include <cmath>3#include <vector>4#include <cstdio>5#include <cstring>6#include <cstdlib>7#include <iostream>8#include <algorithm>9 #defineRep (i,n) for (int i=0;i<n;++i)Ten #defineF (i,j,n) for (int i=j;i<=n;++i) One #defineD (i,j,n) for (int i=j;i>=n;--i) A #definePB Push_back - using namespacestd; - intGetint () { the     intv=0, sign=1;CharCh=GetChar (); -      while(!isdigit (CH)) {if(ch=='-') sign=-1; Ch=GetChar ();} -      while(IsDigit (CH)) {v=v*Ten+ch-'0'; Ch=GetChar ();} -     returnv*Sign ; + } - Const intn=1e7+Ten, inf=~0u>>2; + /*******************template********************/ A  at Doublea[505],b[505],p[ -]; -  - intMain () { - //freopen ("Input.txt", "R", stdin); -     intt=getint (); -     intK1,k2,k3,a,b,c,n; in      while(t--){ -N=getint (); K1=getint (); K2=getint (); k3=getint (); toA=getint (); B=getint (); C=getint (); +         Doublep0=1.0/k1/k2/K3; -Memset (P,0,sizeofp); theF (I,1, K1) F (J,1, K2) F (k,1, K3) *             if(I!=a | | j!=b | | k!=c) p[i+j+k]+=P0; $             Panax NotoginsengD (I,n,0){ -A[i]=p0; b[i]=1; theF (J,1, k1+k2+K3) { +                 if(i+j>n) Break; Aa[i]+=a[i+j]*P[j]; theb[i]+=b[i+j]*P[j]; +             } -         } $printf"%.15lf\n", b[0]/(1-a[0])); $     } -     return 0; -}

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"ZOJ" "3329" one person Game