POJ 3744 Scout YYF I

这题主要是个概率dp，然后加个矩阵快速幂优化
很容易想到dp[i] = dp[i-1]*p+dp[i-2]*(1-p);
然后那个范围是$10^8$所以直接dp会超时的
需要矩阵快速幂优化。
分段去做快速幂
p 1-p
1 0

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <map>

#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define mem(i,a) memset(i,a,sizeof(i))
#define rson mid+1,r,rt<<1|1
#define lson l,mid,rt<<1

#define mp make_pair
#define pb push_back
#define ll long long
#define LL long long
using namespace std;
template <typename T>inline void read(T &_x_){
    _x_=0;bool f=false;char ch=getchar();
    while (ch<'0'||ch>'9') {if (ch=='-') f=!f;ch=getchar();}
    while ('0'<=ch&&ch<='9') {_x_=_x_*10+ch-'0';ch=getchar();}
    if(f) _x_=-_x_;
}
const double eps = 0.0000001;
const int maxn = 3e2+7;
const int mod = 1e9+7;
const ll inf = 1e15;

int n;
double p,ret=1.0;
int pos[15];

struct matrix{
    double m[2][2];
}ans,bs;

matrix multi(matrix a,matrix b){
    matrix c;
    for(int i = 0; i < 2; i++){
        for(int j = 0; j< 2; j++){
            c.m[i][j] = 0;
            for(int k = 0; k< 2; k++){
                c.m[i][j] = (c.m[i][j]+a.m[i][k]*b.m[k][j]);
            }
        }
    }
    return c;
}

void fast_mod(int n){
    bs.m[0][0]=p;bs.m[1][0]=1-p;
    bs.m[0][1]=1;bs.m[1][1]=0;
    ans.m[0][0]=ans.m[1][1]=1;
    ans.m[0][1]=ans.m[1][0]=0;
    while(n){
        if(n&1)
        ans=multi(ans,bs);
        bs=multi(bs,bs);
        n>>=1;
    }
    ret*=(1-ans.m[0][0]);
}

int main(){
    while(~scanf("%d%lf",&n,&p)){
        FOR(i,1,n) scanf("%d",&pos[i]);
        sort(pos+1,pos+1+n);
        ret=1.0;
        for(int i=1;i<=n;i++){
            fast_mod(pos[i]-pos[i-1]-1);
        }
        printf("%.7f\n",ret);
    }
    return 0;
}