Rating
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 714 Accepted Submission(s): 452
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
Author
FZU
Source
Recommend
题意：
一个人打cf。
规则是假设他排名前200就加50分最高加到1000.否側减100分。
最低到0分。
如今告诉你他一场比赛前200的概率p.然后他申请了两个账号初始分都为0.每次比赛他会用分数低的那个账号低的那个账号打。
如今问你他要上1000.有一个账号上即可了。须要參加比赛场数的期望。
思路：
因为加减分都是50的倍数。所以分数能够用[0,20]表示。没次能够减2分或加1分。
用dp[i][j]表示分数高的账号分数为i。分数低的账号分数为j然后有一个账号上1000的概率。
那么dp[i][j]=p*(dp[i][j+1]+1)+(1-p)*(dp[i][j-2]+1) i>j。假设j+1大于i就换成dp[j+1][i]。假设j-2小于0就换成dp[i][0]。然后对每一个dp[i][j]编号。
建立方程。然后高斯消元。
具体见代码：
#include<algorithm>
#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
using namespace std;
const int INF=0x3f3f3f3f;
const double eps=1e-11;
const int maxn=100010;
typedef long long ll;
int cnt,mp[50][50];
double mat[310][310];
bool gauss()
{
int row,i,j,id;
double maxx,var;
for(row=0;row<cnt;row++)
{
maxx=fabs(mat[row][row]);
id=row;
for(i=row+1;i<cnt;i++)//mat[i][cnt]为常数项
{
if(fabs(mat[i][row])>maxx)
{
maxx=fabs(mat[i][row]);
id=i;
}
}
if(maxx<eps)
return false;
if(id!=row)
{
for(i=row;i<=cnt;i++)
swap(mat[row][i],mat[id][i]);
}
for(i=row+1;i<cnt;i++)
{
if(fabs(mat[i][row])<eps)
continue;
var=mat[i][row]/mat[row][row];
for(j=row;j<=cnt;j++)
mat[i][j]-=mat[row][j]*var;
}
}
for(i=cnt-1;i>=0;i--)
{
for(j=i+1;j<cnt;j++)
mat[i][cnt]-=mat[i][j]*mat[j][j];
mat[i][i]=mat[i][cnt]/mat[i][i];
}
return true;
}
int main()
{
int i,j,ptr,base,pp,a,b,c;
double p;
for(i=0;i<=20;i++)
for(j=0,base=i*(i+1)/2;j<=i;j++)
mp[i][j]=base+j;
cnt=231;
while(~scanf("%lf",&p))
{
ptr=0;
memset(mat,0,sizeof mat);
for(i=0;i<=20;i++)
{
for(j=0;j<=i;j++)
{
if(i==20)
{
pp=mp[i][j];
mat[ptr][pp]=1;
mat[ptr++][cnt]=0;
continue;
}
a=max(i,j+1);
b=min(i,j+1);
c=max(0,j-2);
mat[ptr][cnt]=1;
pp=mp[i][j];
mat[ptr][pp]+=1;
pp=mp[a][b];
mat[ptr][pp]+=-p;
pp=mp[i][c];
mat[ptr++][pp]+=p-1;
}
}
gauss();
printf("%.8lf
",mat[0][0]);
}
return 0;
}
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