1) poj-2151-probability dp

Question:

Give you m, t, n, representing t team, do m questions, for all teams, all do one or more questions, and at least one team has the probability of doing n questions.

Practice:

P [I] [j] indicates the probability that the I-th team will answer the j-th question.

Dp [I] [j] [k] represents the I-th team, and the probability that the first j questions will have k questions.

The probability bans * = (1-dp [I] [m] [0]) of all teams on one or more questions // the product of the probability that a team on the other has on the other.

When the probability that all teams have one or more questions:

At least one team has the probability to do n to the question = the probability that all teams have the right to one question-the probability that all teams have the right to less than n questions.

Code:

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# Include <stdio. h>
# Include <iostream>
# Include <string. h>
# Include <algorithm>
# Include <queue>
# Include <stack>
# Include <map>
# Include <string>
# Include <stdlib. h>
# Define INF_MAX 0x7fffffff
# Define INF 999999
# Define max3 (a, B, c) (max (a, B)> c? Max (a, B): c)
# Define min3 (a, B, c) (min (a, B) <c? Min (a, B): c)
# Define mem (a, B) memset (a, B, sizeof ())
Using namespace std;
Struct node
{
Int u;
Int v;
Int w;
Bool friend operator <(node a, node B ){
Return a. w <B. w;
}
} Edge [2, 1001];
Int gcd (int n, int m) {if (n <m) swap (n, m); return n % m = 0? M: gcd (m, n % m );}
Int lcm (int n, int m) {if (n <m) swap (n, m); return n/gcd (n, m) * m ;}
Double p [1, 1001] [51];
Double dp [1, 1001] [31] [31];
Int main ()
{
 
Int m, t, n, I, j, k;
While (scanf ("% d", & m, & t, & n) & (m | n | t ))
{
For (I = 1; I <= t; I ++)
{
For (j = 1; j <= m; j ++)
{
Scanf ("% lf", & p [I] [j]);
}
}
For (I = 1; I <= t; I ++)
{
Dp [I] [0] [0] = 1;
For (j = 1; j <= m; j ++)
{
Dp [I] [0] [j] = 0;
}
}
For (I = 1; I <= t; I ++)
{
For (j = 1; j <= m; j ++)
{
Dp [I] [j] [0] = dp [I] [J-1] [0] * (1-p [I] [j]);
For (k = 1; k <= j; k ++)
{
Dp [I] [j] [k] = dp [I] [J-1] [k] * (1-p [I] [j]) + dp [I] [J-1] [k-1] * p [I] [j];
}
}
}
Double bans; // The probability that all people do the same for one or more questions
Bans = 1;
For (I = 1; I <= t; I ++)
{
Bans * = (1-dp [I] [m] [0]);
}
Double ans; // at least one person has the probability of doing n questions
Ans = 1;
For (I = 1; I <= t; I ++)
{
Double ks;
Ks = 0;
For (j = 1; j <n; j ++)
{
Ks + = dp [I] [m] [j];
}
Ans * = ks;
}
Ans = bans-ans;
Printf ("%. 3f \ n", ans );
}
Return 0;
}

# Include <stdio. h>
# Include <iostream>
# Include <string. h>
# Include <algorithm>
# Include <queue>
# Include <stack>
# Include <map>
# Include <string>
# Include <stdlib. h>
# Define INF_MAX 0x7fffffff
# Define INF 999999
# Define max3 (a, B, c) (max (a, B)> c? Max (a, B): c)
# Define min3 (a, B, c) (min (a, B) <c? Min (a, B): c)
# Define mem (a, B) memset (a, B, sizeof ())
Using namespace std;
Struct node
{
Int u;
Int v;
Int w;
Bool friend operator <(node a, node B ){
Return a. w <B. w;
}
} Edge [2, 1001];
Int gcd (int n, int m) {if (n <m) swap (n, m); return n % m = 0? M: gcd (m, n % m );}
Int lcm (int n, int m) {if (n <m) swap (n, m); return n/gcd (n, m) * m ;}
Double p [1, 1001] [51];
Double dp [1, 1001] [31] [31];
Int main ()
{

Int m, t, n, I, j, k;
While (scanf ("% d", & m, & t, & n) & (m | n | t ))
{
For (I = 1; I <= t; I ++)
{
For (j = 1; j <= m; j ++)
{
Scanf ("% lf", & p [I] [j]);
}
}
For (I = 1; I <= t; I ++)
{
Dp [I] [0] [0] = 1;
For (j = 1; j <= m; j ++)
{
Dp [I] [0] [j] = 0;
}
}
For (I = 1; I <= t; I ++)
{
For (j = 1; j <= m; j ++)
{
Dp [I] [j] [0] = dp [I] [J-1] [0] * (1-p [I] [j]);
For (k = 1; k <= j; k ++)
{
Dp [I] [j] [k] = dp [I] [J-1] [k] * (1-p [I] [j]) + dp [I] [J-1] [k-1] * p [I] [j];
}
}
}
Double bans; // The probability that all people do the same for one or more questions
Bans = 1;
For (I = 1; I <= t; I ++)
{
Bans * = (1-dp [I] [m] [0]);
}
Double ans; // at least one person has the probability of doing n questions
Ans = 1;
For (I = 1; I <= t; I ++)
{
Double ks;
Ks = 0;
For (j = 1; j <n; j ++)
{
Ks + = dp [I] [m] [j];
}
Ans * = ks;
}
Ans = bans-ans;
Printf ("%. 3f \ n", ans );
}
Return 0;
}