HDU 4652 Dice （概率DP）
Dice
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≤10 6) 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 10 9 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
Source SRE实战 互联网时代守护先锋，助力企业售后服务体系运筹帷幄！一键直达领取阿里云限量特价优惠。
field=problem&key=2013%20Multi-University%20Training%20Contest%205&source=1&searchmode=source" style="color:rgb(26,92,200);text-decoration:none;" target="_blank">2013 Multi-University Training Contest 5
题目大意：
m边形的骰子，问你出现连续同样（不同）n次须要掷的次数的数学期望。
解题思路：
利用递归方式的DP的思想推公式
解题代码：（1）若询问为0，则：
dp[i] 记录的是已经连续i个同样，到n个同样同须要的次数的数学期望
dp[0]= 1+dp[1]
dp[1]= 1+( 1/m*dp[2]+(m-1)/m*dp[1])=1+(dp[2]+(m-1)*dp[1])/m;
dp[2]= 1+(dp[3]+(m-1)*dp[2])/m;
....................
dp[n]= 0
推出：
dp[i] = 1 + ( (m-1)*dp[1] + dp[i+1] ) / m
dp[i+1] = 1 + ( (m-1)*dp[1] + dp[i+2] ) / m
因此。m*(dp[i+1]-dp[i])=(dp[i+2]-dp[i+1])
我们发现是等比数列
dp[0]-dp[1]=1;
dp[1]-dp[2]=m;
..........
dp[n-1]-dp[n]=m^(n-1)
累加，得：dp[0]-dp[n]=1+m+m^2+..........m^(n-1)=(1-m^n)/(1-m)
所以：dp[0]=(1-m^n)/(1-m);
（2）若询问为1，则：
dp[0] = 1 + dp[1]
dp[1] = 1 + (dp[1] + (m-1) dp[2]) / m
dp[2] = 1 + (dp[1] + dp[2] + (m-2) dp[3]) / m
dp[i] = 1 + (dp[1] + dp[2] + ... dp[i] + (m-i)*dp[i+1]) / m
dp[i+1]= 1 + (dp[1] + dp[2] + ... dp[i] + dp[i+1] + (m-i-1)*dp[i+1]) / m
...
dp[n] = 0;
选出 dp[i] 和 dp[i+1] 这两行相减 得
dp[i] - dp[i+1] = (m-i-1)/m * (dp[i+1] - dp[i+2]);
因此 dp[i+1] - dp[i+2] = m/(m-i-1)*(dp[i]-dp[i+1]);
所以：
dp[0]-dp[1]=1;
dp[1]-dp[2]=1*m/(m-1);
dp[2]-dp[3]=1*m/(m-1)*m/(m-2);
..........
dp[n-1]-dp[n]=1*m/(m-1)*m/(m-2)*.......*m/(m-n+1);
累加得到答案
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
inline double solve(){
int op,m,n;
scanf("%d%d%d",&op,&m,&n);
double ans=0;
if(op==0){
for(int i=0;i<=n-1;i++){
ans+=pow(1.0*m,i);
}
}else{
double tmp=1.0;
for(int i=1;i<=n;i++){
ans+=tmp;
tmp*=m*1.0/(m-i);
}
}
return ans;
}
int main(){
int t;
while(scanf("%d",&t)!=EOF){
while(t-- >0){
printf( "%.9lf\n",solve() );
}
}
return 0;
}
