# Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 213 Accepted Submission(s): 126

Special Judge

Problem Description
A little girl loves programming competition very much. Recently, she has found a new kind of programming competition named "TopTopTopCoder". Every user who has registered in "TopTopTopCoder" system will have a rating, and the initial
value of rating equals to zero. After the user participates in the contest held by "TopTopTopCoder", her/his rating will be updated depending on her/his rank. Supposing that her/his current rating is X, if her/his rank is between on 1-200 after contest, her/his
rating will be min(X+50,1000). Her/His rating will be max(X-100,0) otherwise. To reach 1000 points as soon as possible, this little girl registered two accounts. She uses the account with less rating in each contest. The possibility of her rank between on
1 - 200 is P for every contest. Can you tell her how many contests she needs to participate in to make one of her account ratings reach 1000 points?
Input
There are several test cases. Each test case is a single line containing a float number P (0.3 <= P <= 1.0). The meaning of P is described above.
Output
You should output a float number for each test case, indicating the expected count of contest she needs to participate in. This problem is special judged. The relative error less than 1e-5 will be accepted.
Sample Input
1.000000
0.814700

Sample Output
39.000000
82.181160


1、能够用高斯消元,

E(X,Y)=PE(X1,Y1)+(1-P)E(X2,Y2)+1,X1,Y1是XY分数赢了的分数。X2,Y2相应是输了的分数。如果X>=Y，每次比赛用Y的账号，就会有210条方程。用mark数组标记XY分数相应的系数的索引。

#include <iostream>
#include <cmath>
#include <stdio.h>
#include <algorithm>
#include <ctime>
#include <vector>
#include <cstring>
#include <map>
#include <string>
#include <queue>
using namespace std;
#define LL long long
#define ULL unsigned long long
//#define REP(i,n) for(int i=0;i<n;++i)
#define REP(i,a,b) for(int i=a;i<=b;++i)
#define INFLL (1LL)<<62
#define mset(a) memset(a,0,sizeof a)
#define FR(a) freopen(a,"r",stdin)
#define FW(a) freopen(a,"w",stdout)
#define PI 3.141592654
const LL MOD = 1000000007;
const int maxn=222;
const double eps=1e-9;
double a[maxn][maxn];
int mark[25][25];
int cnt;
double gauss()
{
int m=211;
int n=210;
for(int i=0;i<n;i++)
{
int k=i;
for(;k<n;++k)
if(fabs(a[k][i])>eps)    break;
if(i!=k)
for(int j=0;j<=n;++j)
swap(a[i][j],a[k][j]);
for(int j=0;j<n;++j)
{
if(i==j)    continue;
if(fabs(a[j][i])<eps)    continue;
double x=a[j][i]/a[i][i];
for(k=i;k<m;++k)
a[j][k]-=a[i][k]*x;
}
}
return a[0][n]/a[0][0];
}

void makeMat(double p)
{
mset(a);
int m=211;
int x=0,y=0;
for(y=0;y<20;++y){
for(x=0;x<y;++x)
{
int temp=mark[y][x];
a[temp][temp]=1;
a[temp][m-1]=1;
int temp2=mark[y][max(0,x-2)];
a[temp][temp2]-=1-p;
temp2=mark[y][x+1];
a[temp][temp2]-=p;
}
int t=mark[y][y];
a[t][t]=1;
a[t][m-1]=1;
int tt=mark[y][max(0,x-2)];
a[t][tt]-=1-p;
tt=mark[x+1][x];
a[t][tt]-=p;
}
}

int main()
{
double p;
cnt=0;
mset(mark);
REP(i,0,20)
REP(j,0,i)
mark[i][j]=cnt++;
while (cin>>p)
{
makeMat(p);
printf("%.6lf\n",gauss());
}
}


2、用dp

dp[i]表示单场比赛从i分数提高到i+1的分数的期望值。

==>dp[i]=1/p+(1-p)/p*(dp[i-2]+dp[i-1]);dp[0]=1/p,dp[1]=1/p/p;

ans[i+1][i]=ans[i][i]+dp[i],ans[i+1][i+1]=ans[i+1][i]+dp[i],

#include <iostream>
#include <cmath>
#include <stdio.h>
using namespace std;
double dp[22];
double ans[22][22];

int main()
{
double p;
while (cin>>p)
{
dp[0]=1/p;
dp[1]=1/p/p;
for(int i=2;i<20;++i)
dp[i] = 1+(1-p)/p*(dp[i-2]+dp[i-1]+1);
ans[0][0]=0;
for (int i=0;i<20;++i)
{
ans[i+1][i]=ans[i][i]+dp[i];
ans[i+1][i+1]=ans[i+1][i]+dp[i];
}
printf("%.6lf\n",ans[20][19]);
}
}


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