"poj3744" Scout yyf I

http://poj.org/problem?id=3744 (Topic link)

Test instructions

Give N Thunder, respectively in A[1]...a[n], take one step probability for P, walk two steps probability for 1-p, a start in position 1th, ask the probability of safe to reach the end.

Solution

It is obvious that dp:f[i]=p*f[i-1]+ (1-p) *f[i-2]. Consider the a[i] position of thunder, then the probability of safe passage is to reach f[a[i]+1] The probability is: f[a[i]-1]* (1-p).

Because A[i] is very large, so we want to segment with a matrix fast power.

Details

The power of the code is down ... Inexplicable wa up can go to see discuss, good pit.

Code

poj3744#include<algorithm> #include <iostream> #include <cstdlib> #include <cstring># include<cstdio> #include <cmath> #define LL long long#define inf 1<<30#define EPS 1e-8#define Pi acos (- 1.0) #define FREE (a) freopen (a ".", "R", stdin), Freopen (a ". Out", "w", stdout); using namespace std;double tmp[2][2],f[2] [2],t[2][2],p;int a[20],n;void Power (int k) {T[0][0]=p,t[0][1]=1,t[1][0]=1-p,t[1][1]=0;while (k) {if (k&1) {for ( int i=0;i<=1;i++) for (int j=0;j<=1;j++) {tmp[i][j]=0;for (int k=0;k<=1;k++) tmp[i][j]+=f[i][k]*t[k][j];} memcpy (F,tmp,sizeof (f)); k>>=1;for (int i=0;i<=1;i++) for (int j=0;j<=1;j++) {tmp[i][j]=0;for (int k=0;k<=1;k++) tmp[i][j]+=t[i][ K]*T[K][J];} memcpy (t,tmp,sizeof (t));}} int main () {while (scanf ("%d%lf", &n,&p)!=eof) {for (int i=1;i<=n;i++) scanf ("%d", &a[i]); Sort (a+1,a+1+n ); a[0]=0;f[1][1]=1;for (int i=1;i<=n;i++) {f[1][0]=f[1][1]*p;f[0][1]=f[1][0]*p+f[1][1]* (1-p); f[0][0]=f[0][1]*p +f[1][0]* (1-p); if (a[i]-a[i-1]==1) {f[1][1]=0;break;} else if (a[i]-a[i-1]==2) f[1][1]=f[1][1]* (1-p), else if (a[i]-a[i-1]==3) f[1][1]=f[1][0]* (1-p), else if (a[i]-a[i-1]==4) f[1][1]=f[0][1]* (1-p); else power (a[i]-a[i-1]-5), f[1][1]= (1-p) *f[0][0];} if (Fabs (f[1][1]) <eps) puts ("0.0000000"), Else printf ("%.7lf\n", F[1][1]);} return 0;}

"poj3744" Scout yyf I