HDU 4870 rating)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 4870

Question: There are two numbers, and the initial score is 0. Each time you choose a game with a small score, if you choose one with the same score, the probability of a P score increases by 50 points, and the maximum score is 1000 points, A probability of 1-P drops by 100,The minimum value is 0.. I would like to ask you the expected number of times you want to play the game when the number reaches 1000.

(X,Y) indicates that the high score is X, and the low score is Y (x> =Y), E (X,Y) indicates from (X,Y) reach (1000,?) Expected number of matches. E (X,Y)=P*E (x1,Y1)+(1-P)*E (X2,Y2)+1, where, (x1,Y1) indicates the status after rating rises. (X2,Y2) indicates the status after rating drops. Every 50 minutes, There are 210 statuses.

E (X,Y)-P*E (x1,Y1 )-(1-P)*E (X2,Y2) =1. There are 210 such equations. X [0] indicates the expectation that E (0, 0) is returned to the target State.

Pay attention to precision.

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL long long#define _LL __int64#define eps 1e-12#define PI acos(-1.0)using namespace std;const int maxn = 231;double a[235][235],p;int equ,var;double x[235];int flag[235][235];int cnt;double Gauss(){    int row,col,max_r;    int i,j;    row = col = 0;    while(row < equ && col < var)    {        max_r = row;        for(i = row+1; i < equ; i++)            if(fabs(a[i][col])-fabs(a[max_r][col]) > eps)                max_r = i;        if(max_r != row)        {            for(j = col; j <= var; j++)                swap(a[row][j],a[max_r][j]);        }        if(fabs(a[row][col]) < eps)        {            col++;            continue;        }        for(i = row+1; i < equ; i++)        {            if(fabs(a[i][col]) > eps)            {                double t = a[i][col]/a[row][col];                a[i][col] = 0.0;                for(j = col+1; j <= var; j++)                    a[i][j] -= a[row][j]*t;            }        }        row++;        col++;    }    for(i = equ-1; i >= 0; i--)    {        if(fabs(a[i][i]) < eps) continue;        double tmp = a[i][var];        for(j = i+1; j < var; j++)            tmp -= a[i][j]*x[j];        x[i] = tmp/a[i][i];    }    return x[0];}int main(){    cnt = 0;    memset(flag,-1,sizeof(flag));    for(int i = 0; i < 20; i++)    {        for(int j = 0; j <= i; j++)            flag[i][j] = cnt++;    }    while(~scanf("%lf",&p))    {        equ = var = 210;        int x;        memset(a,0,sizeof(a));        for(int i = 0; i < 20; i++)        {            for(int j = 0; j < i; j++)            {                x = flag[i][j];                a[x][x] = 1;                a[x][210] = 1;                a[x][flag[i][max(0,j-2)]] -= (1-p);                a[x][flag[i][j+1]] -= p;            }            x = flag[i][i];            a[x][x] = 1;            a[x][210] = 1;            a[x][flag[i][max(0,i-2)]] -= (1-p);            a[x][flag[i+1][i]] -= p;        }        printf("%.6lf\n", Gauss());    }    return 0;}

HDU 4870 rating)