A queue for the first person's situation
Probability P1: The queue remains the same
P2: First person to the end of the team
P3: First person out of the team
P4: System crashes
There are n people in the queue, Tomato in the M position, the system crashes, the number of people in front of the Tomato is less than the probability of the case of the K person
DP[I][J] Indicates the probability that there is an I person in the queue, and that the target State occurs when the Tomato is in position J
You can easily get a recursive formula
DP[I][1] = p2/(1-P1) *dp[i][i] + p4/(1-P1);
DP[I][J] = p2/(1-P1) *dp[i][j-1] + p3/(1-P1) *dp[i-1][j-1] + p4/(1-P1);
DP[I][J] = p2/(1-P1) *dp[i][j-1] + p3/(1-P1) *dp[i-1][j-1];
Only three cases occurred on tomato, successfully out of the team, the system crashes more than K when the system crashes is less than or equal to K
So the initial conditions dp[1][1] = p4/(P3+P4)
And then each time we iterate through the dp[i][i], get the answer.
It is important to note that two-dimensional arrays are hyperspace, so use a scrolling array
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
using namespace Std;
const int MAXN = 2010;
Double DP[2][MAXN];
int n, m, K;
int main ()
{
Freopen ("In.txt", "R", stdin);
Double P1, p2, p3, P4, TMP, temp;
int I, J;
while (~SCANF ("%d%d%d%lf%lf%lf%lf", &n, &m, &k, &p1, &P2, &P3, &P4))
{
memset (DP, 0, sizeof (DP));
if (P4 = = 0)
{
Puts ("0.00000");
Continue;
}
DP[1][1] = p4/(P3+P4);
for (i = 2;i <= n;i++)
{
TMP = 0;
temp = 1;
for (j = i;j > 0; j--)
{
if (j = = 1) tmp+= p4/(1-P1) *temp;
else if (J > K) tmp+=p3/(1-P1) *dp[(i-1)%2][j-1]*temp;
Else tmp+= (p3/(1-P1) *dp[(i-1)%2][j-1] + p4/(1-P1)) *temp;
temp*=p2/(1-P1);
}
Dp[i%2][i] = tmp/(1-temp);
for (j = 1;j < I; j + +)
{
if (j = = 1) dp[i%2][1] = p2/(1-P1) *dp[i%2][i] + p4/(1-P1);
else if (J <= k) dp[i%2][j] = p2/(1-P1) *dp[i%2][j-1] + p3/(1-P1) *dp[(i-1)%2][j-1] + p4/(1-P1);
else dp[i%2][j] = p2/(1-P1) *dp[i%2][j-1] + p3/(1-P1) *dp[(i-1)%2][j-1];
}
}
printf ("%.5lf\n", Dp[n%2][m]);
}
}
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Hdu4089activation probability DP
