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

一场比赛,赢了能够得50分,输了扣100分,分数会超过1000和不会小于0.有个人用2个账号,始终用分数低的号比赛。已知赢一场比赛的概率为p求打到1000分的场数的期望值。

1、能够用高斯消元,

令E(X,Y)为账号分数为x,y打到1000的数学期望。

则有:

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。

代码:

#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

首先离散化,由于每场比赛分数的变化都是50的倍数。令每场赢了得1分。输了扣2分。

dp[i]表示单场比赛从i分数提高到i+1的分数的期望值。

则有:dp[i]=p+(1-p)(dp[i-2]+dp[i-1]+dp[i]+1)

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

用ans[i][i]表示两个账号分数从0打到ii的期望,对于账号分数的上升,他们是交错上升的。意思是当他们分数一样的时候,前面的赢一分,当前面的赢了一分之后,下一场就后面的赢。所以仅仅须要维护ans[i+1][i] 和ans[i+1][i+1],且

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

代码:

#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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