POJ 3744 characteristic equation + fast power

Easy to get: dp[n] = dp[n-1] * p + dp[n-2] * (1-P); (1)

If there is a ray in position I, then: Dp[i + 1] = dp[i-1] * (1-P);

How to obtain Dp[i]?

We can solve the characteristic equation (1) and get:

Dp[n] = A * (p-1) ^ (n-1) + B;

After A and b have been obtained, do a quick power.

We can also get the answer directly with the matrix fast power.

1#include <algorithm>2#include <iostream>3#include <cstring>4#include <cstdio>5 using namespacestd;6 7 Const intN = One;8 intMine[n];9 intN;Ten DoubleA, B, p; One  A DoubleQuickpow (DoubleDintm) - { -     DoubleAns =1.0, W =D; the      while(m) -     { -         if(M &1 ) -         { +Ans = ans *W; -         } +w = w *W; Am = M >>1; at     } -     returnans; - } -  - DoubledpintN) - { in     if(n = =0)return 0; -     returnA * QUICKPOW (P-1N1) +b; to } +  - intMain () the { *      while(SCANF ("%D%LF", &n, &p)! =EOF) $     {Panax NotoginsengA = (P-1.0)/(P-2.0 ); -b =1.0/ (2-p); the          for(inti =0; I < n; i++ ) +         { Ascanf"%d", Mine +i); the         } +Sort (mine, mine +n); -         DoubleAns =1.0; $         intCur =1; $          for(inti =0; I < n; i++ ) -         { -             intdis = mine[i]-cur; theAns = ans * DP (DIS) * (1-p); -Cur = mine[i] +1;Wuyi         } theprintf"%.7f\n", ans); -     } Wu     return 0; -}

POJ 3744 characteristic equation + fast power