POJ 3744 Scout yyf I (Matrix fast Power)

Test instructions: There are n mines on one road, tell you the location of the N thunder, a person initially in position 1, the person can walk 1 units at a time or jump 2 units, the probability of P and 1-p, respectively, now require this person safe passage of the probability of the road

Idea: To find out the probability of safety through each ray, and then all the ride up can be. To safely pass a ray (assuming it is in the place[i] position), only two cells jump from place[i]-1 to pass safely. This only requires that the probability of place[i-1]+1 to Place[i]-1 is multiplied (1-p) to obtain the probability of passing the first ray.

Because of the large data, the matrix must be quickly idempotent. In fact, this is a variant of the Fibonacci sequence: F (i) =f (i-1) *p+f (i-2) * (1-p); to quickly find F (i), use the matrix to quickly power.

#include <iostream>#include<cstdio>#include<algorithm>#include<cstring>using namespacestd;intplace[1010] ;Doubleep[1010], p, F1, F2;structmat{Doublemat[Ten][Ten] ;};    Mat Mul (Mat A,mat b) {mat ret;  for(intI=0;i<2; i++)       for(intj=0;j<2; j + +) {Ret.mat[i][j]=0;  for(intk=0;k<2; k++) Ret.mat[i][j]+=a.mat[i][k]*B.mat[k][j]; }    returnret;} Mat Pow_m (Mat A,intN)    {MAT ret; memset (Ret.mat,0,sizeof(Ret.mat));  for(intI=0;i<2; i++) ret.mat[i][i]=1; Mat Temp=A;  while(n) {if(n&1) ret=Mul (ret,temp); Temp=Mul (temp,temp); N>>=1; }    returnret;}DoubleQ_mul (intk)    {Mat A; a.mat[0][0]=p, a.mat[0][1]=1.0-P, a.mat[1][0]=1.0, a.mat[1][1]=0; MAT ret=pow_m (a,k); return(f2*ret.mat[1][0]+f1*ret.mat[1][1])*(1.0-p);}intMain () {intN;  while(cin>>n>>p) {memset (place,0,sizeof(place));  for(intI=0; i<n;i++) {scanf ("%d",&Place[i]); } sort (Place,place+N); intnow=1; Doubleans=1.0; if(place[0]==1) {printf ("0.0000000\n") ; Continue ; }         for(intI=0; I<n; i++){            //cout<<i<< "<<place[i]<<" "<<now<<" "<<endl;            if(Place[i] = =Now ) { Now= -1;  Break; }            Else{F1=1.0; if(Now +1==Place[i]) {                    //cout<<1<<endl;Ep[i] = f1* (1.0-p); }                Else{F2=f1* (1.0*p); Ep[i]=q_mul (place[i]-(now+1)) ; //cout<<place[i]-(now+1) <<endl;} Now=place[i] +1 ; }        }        if(now==-1) {printf ("0.0000000\n") ; }        Else{             for(intI=0; I <n; i++){                //cout<<ep[i]<<endl;ans*=Ep[i]; } printf ("%.7f\n", ans); }     }    return 0;}

POJ 3744 Scout yyf I (Matrix fast Power)