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) + qsum_{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) + qsum_{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 << "
";
#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
", ans);
    }
    return 0;
}