Kejin Player （概率DP）hdu6656

原题目：

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 ii , you have to pay aiai RMB to get to level i+1i+1 ". Cuber QQ now changed it a little bit: "when you are level ii , you pay aiai RMB, are you get to level i+1i+1 with probability pipi ; otherwise you will turn into level xixi (xi≤ixi≤i )".

Cuber QQ still needs to know how much money expected the Kejin players needs to ``ke'' so that they can upgrade from level ll to level rr , 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 tt , denoting the number of test cases.

For each test case, there is two space-separated integers nn (1≤n≤500 0001≤n≤500 000 ) and qq (1≤q≤500 0001≤q≤500 000 ) in the first line, meaning the total number of levels and the number of queries.

Then follows nn lines, each containing integers riri , sisi , xixi , aiai (1≤ri≤si≤1091≤ri≤si≤109 , 1≤xi≤i1≤xi≤i , 0≤ai≤1090≤ai≤109 ), space separated. Note that pipi is given in the form of a fraction risirisi .

The next qq lines are qq queries. Each of these queries are two space-separated integers ll and rr (1≤l<r≤n+11≤l<r≤n+1 ).

The sum of nn and sum of qq from all tt test cases both does not exceed 106106 .

Output

For each query, output answer in the fraction form modulo 109+7109+7 , that is, if the answer is PQPQ , you should output P⋅Q−1P⋅Q−1 modulo 109+7109+7 , where Q−1Q−1 denotes the multiplicative inverse of QQ modulo 109+7109+7 .

Sample Input

Sample Output

22
12

中文概要：
（游戏升级）共有n+1个级别，给出前n个级别的属性，分别为r,s,x,a( r/s ：成功升至第i+1级的概率，x ：若升级不成功，则掉至第x级，且x比当前 i 小，a：由当前级升至第i+1级时所需要的花费)。现给出q个询问：每次求从第 L 级升至 R 级需要钱的期望。

#include <bits/stdc++.h>
using namespace std;
const int modd=1e9+7;
const int maxn=500005;
long long dp[500005];
long long qkpow(long long a,long long p,long long m)
{    long long t=1,tt=a%m;    while(p)   {        if(p&1)t=t*tt%m;        tt=tt*tt%m;        p>>=1;    }    return t;}//快速幂求逆元long long getInv2(long long a,long long mod){ return qkpow(a,mod-2,mod);}int main()
{int t;int n,q,r,s,x,a;long long tmp;scanf("%d",&t);while(t--){scanf("%d%d",&n,&q);dp[1]=0;for(int i=1;i<=n;i++){scanf("%d%d%d%d",&r,&s,&x,&a);dp[i+1]=(dp[i]+a+modd)%modd;//防止得到负数dp[i+1]=(dp[i+1]+(r-s)%modd*getInv2(s,modd)%modd*dp[x]%modd+modd)%modd;dp[i+1]=(dp[i+1]%modd*s%modd*getInv2(r,modd)%modd)%modd;}int xx,yy;       for(int i=0; i<q; i++)        {            scanf("%d%d",&xx,&yy);            printf("%lld\n",(dp[yy]-dp[xx]+modd)%modd);       }    }    return 0;
}

思路：

因为只能一级一级的往上升，所以存在递推关系：（队内大佬教的）

dp[i+1] = (r/s)*(dp[i] + a)+(1 - r/s)*( dp[i]+a+dp[i+1]-dp[x] )

(前者为成功需要的钱，后者为不成功所需要的钱)

=> dp[i+1] = dp[i] + a + (1 - r/s)*( dp[i+1] - dp[x] )

(dp[i]表示升至第i级所需要的金钱期望)

化简得 => dp[i+1] = (s/r)*( dp[i] + a - dp[x] ) + dp[x+1]

注意：该题给的数很大，在每次算加法时都要防止所得的数为负数，故要（+modd）%modd

递推式还好，但是那个逆元还是不太会。。。

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