Coins I
时间限制: 1 Sec 内存限制: 128 MB
提交: 101 解决: 60
[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题意：有n枚朝下的硬币，我们可以投掷这些硬币m次，每次投掷 t 枚硬币，问最后朝上硬币的期望
答安就是有0到n枚硬币朝上，每种情况的概率*相应情况的加和，0*概率[0]+1*概率[1]+...
dp[i][j]代表前i次投掷正面向上的个数为j的概率
代码：
#include<bits/stdc++.h>
using namespace std;
double c[103][103],p[103],dp[103][103];
int main()
{
int n,m,k;
int t; c[0][0]=1;
for(int i=1; i<=100; i++)
{
c[i][0]=1;
for(int j=1; j<=i; j++)
{
c[i][j]=c[i-1][j-1]+c[i-1][j];
}
}
p[0]=1;
for(int i=1; i<=100; i++)
{
p[i]=p[i-1]/2.0;
}
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&k);
memset(dp,0,sizeof(dp));
dp[1][0]=1;
for(int i=1; i<=m+1; i++)
{
for(int j=0; j<=n; j++)
{
for(int l=0; l<=k; l++)//下次投掷会增加的数量
{
if(n-j>=k)//剩余反面的硬币数大于等于k个
{
dp[i+1][j+l]+=dp[i][j]*c[k][l]*p[k];
}
else
{
dp[i+1][j-(k-(n-j))+l]+=dp[i][j]*c[k][l]*p[k];
//剩余反面个数小于k个，那就将状态转移给在原来j个正面的基础上减去已经是正面的个数，因为此时才是可以下一次投掷增加l个也就是0-k个
}
}
}
}
double ans=0;
for(int i=1; i<=n; i++)
{
ans+=dp[m+1][i]*(i*1.0);//将相应概率和相应情况相乘
}
printf("%.3f\n",ans);
}
}