HDU6656 Kejin Player

Kejin Player Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 1082    Accepted Submission(s): 423   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 ai RMB to get to level i+1 ". Cuber QQ now changed it a little bit: "when you are level i , you pay ai RMB, are you get to level i+1 with probability pi ; otherwise you will turn into level xi (xi≤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≤n≤500 000 ) and q (1≤q≤500 000 ) in the first line, meaning the total number of levels and the number of queries. Then follows n lines, each containing integers ri , si , xi , ai (1≤ri≤si≤109 , 1≤xi≤i , 0≤ai≤109 ), space separated. Note that pi is given in the form of a fraction risi . The next q lines are q queries. Each of these queries are two space-separated integers l and r (1≤l The sum of n and sum of q from all t test cases both does not exceed 106 .     Output For each query, output answer in the fraction form modulo 109+7 , that is, if the answer is PQ , you should output P⋅Q−1 modulo 109+7 , where Q−1 denotes the multiplicative inverse of Q modulo 109+7 .   Sample Input 1 3 2 1 1 1 2 1 2 1 3 1 3 3 4 1 4 3 4 Sample Output 22 12 HintHuge IO (Over 40MB)! IO optimization is preferred.

题意：（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。

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，在这里我得到了惨痛的狂wa

并且此题涉及到（a/b）%modd 故要要用到求逆元算法（详见此处  悄悄推销自己哈哈哈）

代码：

#include 
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()
{
    //freopen("in.txt","r",stdin);
    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