5212: Coins I
时间限制: 1 Sec 内存限制: 128 MB
提交: 100 解决: 59
[提交] [状态] [讨论版] [命题人:admin]
题目描述
Alice and Bob are playing a simple game. They line up a row of n identical coins, all with the heads facing down onto the table and the tails upward.
For exactly m times they select any k of the coins and toss them into the air, replacing each of them either heads-up or heads-down with the same possibility. Their purpose is to gain as many coins heads-up as they can.
输入
The input has several test cases and the first line contains the integer t (1 ≤ t ≤ 1000) which is the total number of cases.
For each case, a line contains three space-separated integers n, m (1 ≤ n, m ≤ 100) and k (1 ≤ k ≤ n).
输出
For each test case, output the expected number of coins heads-up which you could have at the end under the optimal strategy, as a real number with the precision of 3 digits.
样例输入
6 2 1 1 2 3 1 5 4 3 6 2 3 6 100 1 6 100 2
样例输出
0.500 1.250 3.479 3.000 5.500 5.000
来源/分类
【题意】
桌面上有n个硬币全部反着，现在你有m次操作机会，每次可以任选k个，然后抛向空中，每个落下来有0.5的概率正，0.5反。
问，在尽量使硬币为正的策略下，最后的正面期望值。
【分析】
dp[i][j]表示第i次操作，正面朝上的有j个的概率。
枚举次数，枚举正面数，枚举本次要把几个抛正。
【代码】
#include<bits/stdc++.h>
using namespace std;
double two[120];
double C[110][110];
double dp[110][110];
int main()
{
C[0][0]=1;//特殊
for(int i=1;i<105;i++)
for(int j=0;j<=i;j++)
C[i][j]=(j == 0) ? 1:(C[i-1][j]+C[i-1][j-1]);
two[0]=1.0;
for(int i=1;i<=100;i++) two[i]=two[i-1]*0.5;
int T,n,m,k;
cin>>T;
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
for(int i=0;i<=m;i++)
for(int j=0;j<=n;j++)
dp[i][j]=0;
dp[0][0]=1.0;
for(int i=0;i<m;i++)
{
for(int j=0;j<=n;j++) //枚举up
{
for(int x=0;x<=k;x++) //枚举我要反过来x个,最多k个
{
double p=C[k][x]*two[k];//从k个中选x个抛成正的概率
if(n-j>=k) //直接反所有down的
dp[i+1][j+x]+=dp[i][j]*p;
else
dp[i+1][n-k+x]+=dp[i][j]*p; //反的全选，再选一部分正的，用原来正n-k加上现在弄正的x个0
}
}
}
double ans=0;
for(int i=0;i<=n;i++)
ans+=dp[m][i]*i;
printf("%.3f\n",ans);
}
}