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  • codeforce Gym 100425E The Street Escalator(期望,线性递推)

    算数学期望,每个人都可以分开来考虑。Xi表示第i个人跑到另外一边的次数。

    Xi服从二项分布。概率的和是个二项式,(p+1-p)^T,把二项式展开,p的偶次项是留在原来那一边的概率。

    可以用((a+b)^T+(a-b)^T)/2来算出偶次项之和。

    也可以用矩阵快速幂。矩阵构造如下

    #include<cstdio>
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<vector>
    #include<map>
    #include<set>
    #include<algorithm>
    #include<cmath>
    //#include<bits/stdc++.h>
    using namespace std;
    
    typedef double MType;
    const int maxn = 1e5+5;
    
    struct Person
    {
        double p;
        char side;
        int T;
        void IN(){
            scanf("%d %c %lf",&T,&side,&p);
        }
        void cal(MType &lft, MType &rgh){
            double keep = (1+pow(1-2*p,T))/2;
            if(side == 'L') lft = keep, rgh = 1-keep;
            else lft = 1-keep, rgh = keep;
        }
    }P;
    
    //#define LOCAL
    int main()
    {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
        int N; scanf("%d",&N);
        MType L = 1, R = 0, ans = 0;
        for(int i = 0; i < N; i++){
            P.IN();
            MType l,r;
            P.cal(l,r);
            ans += L*l + R*r;
            L = r;
            R = l;
        }
        printf("%.12lf",ans);
        return 0;
    }

    快速幂

    #include<cstdio>
    #include<iostream>
    #include<string>
    #include<cstring>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<vector>
    #include<map>
    #include<set>
    #include<algorithm>
    #include<cmath>
    //#include<bits/stdc++.h>
    using namespace std;
    
    const int MSIZE = 2, n = 2;
    typedef double MType;
    struct Matrix
    {
        MType dat[MSIZE][MSIZE];
        MType *operator [](int x){ return dat[x]; }
        Matrix operator * (Matrix& B) {
            Matrix re;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < n; j++){
                    re[i][j] = 0;
                    for(int k = 0; k < n; k++){
                        re[i][j] += dat[i][k]*B[k][j];
                    }
                }
            }
            return re;
        }
        Matrix operator ^ (int q){
            Matrix Re, A = *this;
            for(int i = 0; i < n; i++){
                for(int j = 0; j < n; j++){
                    Re[i][j] = i == j?1:0;
                }
            }
            while(q){
                if(q&1){
                    Re = Re * A;
                }
                A = A * A;
                q >>= 1;
            }
            return Re;
        }
    };
    
    const int maxn = 1e5+5;
    struct Person
    {
        double p;
        char side;
        int T;
        void IN(){
            scanf("%d %c %lf",&T,&side,&p);
        }
        void cal(MType &lft, MType &rgh){
            Matrix C;
            C[0][0] = C[1][1] = 1-p;
            C[0][1] = C[1][0] = p;
            C = C^T;
            if(side == 'L') lft = C[0][0], rgh = C[1][0];
            else lft = C[0][1], rgh = C[1][1];
        }
    }P[maxn];
    
    //#define LOCAL
    int main()
    {
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
        int N; scanf("%d",&N);
        for(int i = 0; i < N; i++){
            P[i].IN();
        }
        //sort(P,P+N);
        MType L = 1, R = 0, ans = 0;
        for(int i = 0; i < N; i++){
            MType l,r;
            P[i].cal(l,r);
            ans += L*l + R*r;
            L = r;
            R = l;
        }
        printf("%.12lf",ans);
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jerryRey/p/4890122.html
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