POJ 3744 水概率dp+水矩陣快速冪

思路：很水的概率dp，很水的矩陣快速冪，就是從當前點推出從當前點到後兩個點的概率，然後因爲n比較大，所以用矩陣快速冪優化（不過我寫的這個還是跑得慢= =

代碼：

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
using namespace std;
const int N = 130;
const int mod = 1e9+7;
#define ll long long
struct Matrix{
    double a[N][N];
    Matrix(){
        memset(a,0,sizeof(a));
        for(int i = 0;i < N;i ++)
            a[i][i] = 1.0;
    }
};
int n = 2;
Matrix multiply(Matrix a,Matrix b){
    Matrix ans;
    memset(ans.a,0,sizeof(ans.a));
    for(int i = 0;i < n;i ++){
        for(int j = 0;j < n;j ++){
            if(a.a[i][j] > 1e-10){
                for(int k = 0;k < n;k ++){
                    ans.a[i][k] = (ans.a[i][k] + a.a[i][j] * b.a[j][k]);//%mod)%mod;
                }
            }
        }
    }
    return ans;
}
Matrix add(Matrix a,Matrix b){
    Matrix ans;
    for(int i = 0;i < n;i ++)
        for(int j = 0;j < n;j ++){
            ans.a[i][j] = (a.a[i][j] + b.a[i][j]);//%mod;
        }
    return ans;
}
Matrix fast_mod(Matrix A,ll k){
    Matrix ans;
    while(k){
        if(k&1)ans = multiply(ans,A);
        A = multiply(A,A);
        k = k>>1;
    }
    return ans;
}
int main(){
    int nn;
    double p,q;
    while(~scanf("%d%lf",&nn,&p)){
        q = 1-p;
        Matrix now;
        now.a[0][0] = 0;
        now.a[0][1] = 1;
        now.a[1][0] = q;
        now.a[1][1] = p;
        double b = 1;
        int last = 0;
        bool flag = 0;
        int num[20];
        for(int i = 0;i < nn;i ++)
            scanf("%d",&num[i]);
        sort(num,num+nn);
        for(int i = 0;i < nn;i ++){
            int tt = num[i]; //cout<<"aaaaaa"<<tt-last-1<<endl;
            if(tt == last)continue;
            if(tt - last == 1){flag = 1;break;}
            Matrix tmp = fast_mod(now,tt - last - 1);
            b = tmp.a[0][1]*b;
            b = b*q;
            last = tt;
        }
        if(flag == 1)b = 0;
        printf("%.7f\n",b);
    }
    return 0;
}