Sgu495: probability DP/push Formula

Probability question .. You can use either DP or formula.

Abstract questions:

There are n small balls, and the expected number of retrieved small balls is retrieved m times.

DP maintains the two States. The first retrieved is the probability of a ball that has not been taken out. DP [I] and the probability of a ball that has been taken out is NP [I];

If the second I-1 is taken out of the ball has been taken out, then the probability that the second I was not taken out of the ball is DP [I-1];

On the contrary, it is DP [I-1]-1/N (the ball that has not been taken out is missing one)

So we can obtain the state transition equation DP [I] = DP [I-1] * (DP [I-1]-1/n) + NP [I-1] * DP [I-1];

You can also push the formula .. However, I still think that the formula is based on the character. It is of course excellent to be YY-ready...

Code:

 

#include <iostream>#include <stdio.h>#include<string.h>#include<algorithm>#include<string>#include<math.h>#include<ctype.h>using namespace std;#define MAXN 10000int n,m;double dp[100010];double np[100010];double solve(){    double res=0;    memset(dp,0,sizeof(dp));    memset(np,0,sizeof(np));    dp[1]=1;    np[1]=0;    for(int i=2;i<=m;i++)    {        dp[i]=dp[i-1]*(dp[i-1]-1.0/(double)n)+np[i-1]*dp[i-1];        np[i]=1-dp[i];    }    for(int i=1;i<=m;i++)    {        res+=dp[i];    }    return res;}int main(){    while(scanf("%d%d",&n,&m)!=EOF)    {        printf("%.10lf\n",solve());    }    return 0;}

Formula ..

#include <stdio.h>#include<math.h>double n,m;int main(){    while(scanf("%lf%lf",&n,&m)!=EOF)    {        printf("%.10lf\n",n-n*pow(((n-1)/n),m));    }    return 0;}

 

Sgu495: probability DP/push Formula