HDU 6656 Kejin Player (期望DP 逆元)

2019 杭电多校 7 1011

题目链接：HDU 6656

比赛链接：2019 Multi-University Training Contest 7

Problem Description

Cuber QQ always envies those Kejin players, who pay a lot of RMB to get a higher level in the game. So he worked so hard that you are now the game designer of this game. He decided to annoy these Kejin players a little bit, and give them the lesson that RMB does not always work.

This game follows a traditional Kejin rule of "when you are level i, you have to pay $a_i$ RMB to get to level $i+1$". Cuber QQ now changed it a little bit: "when you are level $i$, you pay $a_i$ RMB, are you get to level $i+1$ with probability $p_i$; otherwise you will turn into level $x_i (x_i\le i)$".

Cuber QQ still needs to know how much money expected the Kejin players needs to ``ke'' so that they can upgrade from level $l$ to level $r$, because you worry if this is too high, these players might just quit and never return again.

Input

The first line of the input is an integer t, denoting the number of test cases.

For each test case, there is two space-separated integers $n (1\le n\le 500 000)$ and $q (1\le q\le 500 000)$ in the first line, meaning the total number of levels and the number of queries.

Then follows $n$ lines, each containing integers $r_i, s_i, x_i, a_i$ $(1\le r_i\le s_i\le 10^9, 1\le x_i\le i, 0\le a_i\le 10^9)$, space separated. Note that $p_i$ is given in the form of a fraction $\frac{r_i}{s_i}$.

The next $q$ lines are $q$ queries. Each of these queries are two space-separated integers $l$ and $r$ $(1\le l < r\le n+1)$.

The sum of $n$ and sum of $q$ from all $t$ test cases both does not exceed $10^6$.

Output

For each query, output answer in the fraction form modulo $10^9+7$, that is, if the answer is $\frac{P}{Q}$, you should output $P\cdot Q^{−1}$ modulo $10^9+7$, where $Q^{−1}$ denotes the multiplicative inverse of $Q$ modulo $10^9+7$.

Sample Input

Sample Output

22

12

Solution

题意：

从 $i$ 级升级到 $i + 1$ 级需要花费 $a_i$ RMB，成功的概率为 $p_i = \frac{r_i}{s_i}$，若失败则降到 $x_i$ 级，然后给出 $q$ 个询问求 $l$ 级升级到 $r$ 级花费的期望。

题解：

期望DP 逆元

设 $g(l, r)$ 为 $l$ 升到 $r$ 的期望，这种期望满足减法 $g(l, r) = g(1, r) − g(1, l)$。因为升级只能一级一级升, 所以要从 $1$ 升级到 $r$, 必然要经过 $l$。可以降维，用 $dp[i]$ 表示从 $1$ 升到 $i$ 的期望，则 $g(l, r) = dp[r] − dp[l]$。

从 $dp[i]$ 转移至 $dp[i + 1]$，假设尝试了 $t$ 次才成功，那么也就是前面 $t - 1$ 次都是失败的，所以下一状态的花费为当前状态的花费 + 成功的花费 + 失败的花费 + 失败后再次回到当前状态的花费。于是：

\[dp[i + 1] = dp[i] + 1 \times a[i] + (t - 1) \times a[i] + (t- 1) \times (dp[i] - dp[x_i])
\]

又 $\frac{t - 1}{t} = 1 - \frac{r_i}{s_i}$，即 $t = \frac{s_i}{r_i}$

于是状态转移方程为：

\[dp[i + 1] = dp[i] + \frac{s_i}{r_i} \times a[i] + (\frac{s_i}{r_i} - 1) \times (dp[i] - dp[x_i])
\]

Code

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 5e5 + 10;

const ll mod = 1e9 + 7;

ll r[maxn], s[maxn], x[maxn], a[maxn];

ll dp[maxn];

ll qmod(ll a, ll b, ll p) {

    ll ans = 1;

    while(b) {

        if(b & 1) ans = (a * ans) % p;

        a = (a * a) % p;

        b >>= 1;

    }

    return ans;

}

int main() {

    int T;

    cin >> T;

    while(T--) {

        int n, q;

        scanf("%d%d", &n, &q);

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

            scanf("%lld%lld%lld%lld", &r[i], &s[i], &x[i], &a[i]);

            ll t = (s[i] * qmod(r[i], mod - 2, mod)) % mod;

            dp[i + 1] = (dp[i] + (t * a[i]) % mod + ((t - 1) * (dp[i] - dp[x[i]])) % mod + mod) % mod;

        }

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

            int l, r;

            scanf("%d%d", &l, &r);

            printf("%lld\n", (dp[r] - dp[l] + mod) % mod);

        }

    }

    return 0;

}