hdu多校第七场 1011 （hdu6656） Kejin Player 概率dp

题意：

一个游戏，有许多关，到下一关要花费金钱，做出尝试，有概率成功，若成功则到达下一关，若失败则停在此关或退回到前面某关，询问第l关到第r关的期望费用

题解：

显然，第r关到第l关的费用是dp[r]-dp[l]

那么如何算出dp数组呢？首先dp[1]=0，利用期望方程正推

假设i点，成功率为p，失败则跳到j，成功则跳到k，花费q，则期望方程为

dp[k]=dp[i]+p(q)+(1-p)(dp[k]-dp[j])

移项后递推即可

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long long ll;
typedef pair<int, LL>P;
const int M = 4e5 *  + ;
const LL mod = 1e9 + ;
const LL lINF = 0x3f3f3f3f3f3f3f3f;
#define ls (rt<<1)
#define rs (rt<<1|1)
LL gcd(LL a, LL b) { return b ? gcd(b, a%b) : a; }
LL quickpow(LL a, LL b, LL mod)
{
    LL res = ;
    while (b)
    {
        if (b & )
            res = (res*a) % mod;
        a = (a*a) % mod;
        b >>= ;
    }
    return res;
}
inline int read()
{
    int x = , f = ;
    char ch = getchar();
    while (ch < '' || ch > '')
    {
        if (ch == '-')
            f = -;
        ch = getchar();
    }
    while (ch >= '' && ch <= '')
    {
        x =  * x + ch - '';
        ch = getchar();
    }
    return x * f;
}
inline ll readll()
{
    ll x = , f = ;
    char ch = getchar();
    while (ch < '' || ch > '')
    {
        if (ch == '-')
            f = -;
        ch = getchar();
    }
    while (ch >= '' && ch <= '')
    {
        x =  * x + ch - '';
        ch = getchar();
    }
    return x * f;
}
int n, q;
LL r[M], s[M], x[M], a[M], dp[M];
int main()
{
    int t = read();
    while (t--)
    {
        n = readll();
        q = readll();
        for (int i = ; i <= n; i++)
        {
            r[i] = readll();
            s[i] = readll();
            x[i] = readll();
            a[i] = readll();
        }
        dp[] = quickpow(r[], mod - , mod)*a[] % mod*s[] % mod;
        for (int i = ; i <= n; i++)
        {
            LL tmp = ((a[i] + (s[i] - r[i])*quickpow(s[i], mod - , mod) % mod*(dp[i] - dp[x[i]] + mod) % mod) % mod) % mod*s[i] % mod * quickpow(r[i], mod - , mod) % mod;
            dp[i + ] = (dp[i] + tmp) % mod;
        }
        while (q--)
        {
            int l, r;
            l = read();
            r = read();
            printf("%lld\n", (dp[r] - dp[l] + mod) % mod);
        }
    }
}