HDU 4870

Rating

高斯消元。变量不多。

Rating

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 166 Accepted Submission(s): 87 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

2014 Multi-University Training Contest 1

其实变量数很少，直接高斯消元。   其中在求解过程判断了 < eps 返回无解，WA了好多发。 注释掉就AC了。

/* ***************

Author        :kuangbin

Created Time  :2014/7/22 23:33:49

File Name     :E:\2014ACM\比赛\2014多校训练\2014多校1\HDU4870.cpp

************************************************ */

#include <stdio.h>

#include <string.h>

#include <iostream>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <time.h>

using namespace std;

#define eps 1e-9

const int MAXN=420;

double a[MAXN][MAXN],x[MAXN];//方程的左边的矩阵和等式右边的值，求解之后x存的就是结果

int equ,var;//方程数和未知数个数

/*

*返回0表示无解，1表示有解

*/

int Gauss()

{

    int i,j,k,col,max_r;

    for(k=0,col=0;k<equ&&col<var;k++,col++)

    {

        max_r=k;

        for(i=k+1;i<equ;i++)

          if(fabs(a[i][col])>fabs(a[max_r][col]))

            max_r=i;

      //  if(fabs(a[max_r][col])<eps)return 0;

        if(k!=max_r)

        {

            for(j=col;j<var;j++)

              swap(a[k][j],a[max_r][j]);

            swap(x[k],x[max_r]);

        }

        x[k]/=a[k][col];

        for(j=col+1;j<var;j++)a[k][j]/=a[k][col];

        a[k][col]=1;

        for(i=0;i<equ;i++)

          if(i!=k)

          {

              x[i]-=x[k]*a[i][col];

              for(j=col+1;j<var;j++)a[i][j]-=a[k][j]*a[i][col];

              a[i][col]=0;

          }

    }

    return 1;

}

int id[22][22];

int main()

{

    //freopen("1010.in","r",stdin);

    //freopen("out.txt","w",stdout);

	int cnt = 0;

	for(int i = 0;i <= 20;i++)

		for(int j = 0;j <= 20;j++)

		{

			if(i > j)continue;

			if(i == 20 && j == 20)continue;

			id[i][j] = cnt++;

		}

	equ = var = cnt;

	double p;

	while(scanf("%lf",&p) == 1)

	{

		memset(a,0,sizeof(a));

		for(int i = 0;i <= 20;i++)

			for(int j = 0;j <= 20;j++)

			{

				if(i > j)continue;

				if(i == 20 && j == 20)continue;

				int u = id[i][j];

				a[u][u] = 1.0;

				if(i == 20 || j == 20)

				{

					x[u] = 0.0;

					continue;

				}

				x[u] = 1.0;

				int nx = i+1;

				if(nx <= j)

					a[u][id[nx][j]] -= p;

				else a[u][id[j][nx]] -= p;

				nx = max(0,i-2);

				a[u][id[nx][j]] -= (1-p);

			}

		Gauss();

		printf("%.6lf\n",x[0]);

	}

    return 0;

}