HDU 6568 Math

Math

\(f_i\)为从\(i\)到\(i+1\)的期望步数。

\(f_i = 1-p + p(f_i + 2((1-q)^{n-i}(n-i) + q\sum_{j=0}^{n-i-1}(1-q)^{j}j))\)

移项相减得：

\(f_i = 1+\frac{2p((1-q)^{n-i}(n-i) + q\sum_{j=0}^{n-i-1}(1-q)^{j}j)}{1-p}\)

然后预处理一个前缀和就可以了。

代码：

#pragma GCC optimize(2)

#pragma GCC optimize(3)

#pragma GCC optimize(4)

#include<bits/stdc++.h>

using namespace std;

#define y1 y11

#define fi first

#define se second

#define pi acos(-1.0)

#define LL long long

//#define mp make_pair

#define pb emplace_back

#define ls rt<<1, l, m

#define rs rt<<1|1, m+1, r

#define ULL unsigned LL

#define pll pair<LL, LL>

#define pli pair<LL, int>

#define pii pair<int, int>

#define piii pair<pii, int>

#define pdd pair<double, double>

#define mem(a, b) memset(a, b, sizeof(a))

#define debug(x) cerr << #x << " = " << x << "\n";

#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

//head

const int N = 1e5 + 10;

int n;

double p, q, _q[N], sum[N];

int main() {

    while(~scanf("%d %lf %lf", &n, &p, &q)) {

        sum[0] = 0;

        _q[0] = 1;

        for (int i = 1; i <= n; ++i) {

            _q[i] = _q[i-1]*(1-q);

            sum[i] = sum[i-1] + _q[i]*i;

        }

        double ans = 0;

        for (int i = 0; i < n; ++i) {

            ans += 1+(2*p*(_q[n-i]*(n-i)+q*sum[n-i-1]))/(1-p);

        }

        printf("%.10f\n", ans);

    }

    return 0;

}

巴特西

HDU 6568 Math

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