程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU 4652 Dice（概率dp）

HDU 4652 Dice（概率dp）

編輯：C++入門知識

HDU 4652 Dice（概率dp）

Problem Description You have a dice with m faces, each face contains a distinct number. We assume when we tossing the dice, each face will occur randomly and uniformly. Now you have T query to answer, each query has one of the following form:
0 m n: ask for the expected number of tosses until the last n times results are all same.
1 m n: ask for the expected number of tosses until the last n consecutive results are pairwise different.
Input The first line contains a number T.(1≤T≤100) The next T line each line contains a query as we mentioned above. (1≤m,n≤10⁶) For second kind query, we guarantee n≤m. And in order to avoid potential precision issue, we guarantee the result for our query will not exceeding 10⁹ in this problem.
Output For each query, output the corresponding result. The answer will be considered correct if the absolute or relative error doesn't exceed 10^-6.
Sample Input

6
0 6 1
0 6 3
0 6 5
1 6 2
1 6 4
1 6 6
10
1 4534 25
1 1232 24
1 3213 15
1 4343 24
1 4343 9
1 65467 123
1 43434 100
1 34344 9
1 10001 15
1 1000000 2000

Sample Output

1.000000000
43.000000000
1555.000000000
2.200000000
7.600000000
83.200000000
25.586315824
26.015990037
15.176341160
24.541045769
9.027721917
127.908330426
103.975455253
9.003495515
15.056204472
4731.706620396

參考別人的博客：http://blog.csdn.net/auto_ac/article/details/9919851

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define eps 1e-8
typedef __int64 ll;

#define fre(i,a,b)  for(i = a; i = a;i--)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define ssf(n)      scanf("%s", n)
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 10005

int n,m;
double pp;

void solve()
{
	int i,j;
    double ans=0;
   	fre(i,0,n)
   	  ans+=pow(pp,i);
	pf("%.7f\n",ans);
}

void solvee()
{
	 int i,j;
	 double t=1;

	 double ans=0;
	 double up=m,down=m;
	 double temp=up/down;
	 fre(i,0,n)
	 {
	 	ans+=temp;
	 	down--;
	 	temp*=up/down;
	 }
    pf("%.7f\n",ans);
}

int main()
{
    int i,j,t;
    while(~sf(t))
	{
		int op;
		while(t--)
		{
			sfff(op,m,n);
			if(op==0)
				{
					pp=m;
				   solve();
				}
			else
				{
					solvee();
				}
		}
	}
  return 0;
}