habib_rahman's blog

By habib_rahman, history, 5 years ago, In English

Problem:

As Harry Potter series is over, Harry has no job. Since he wants to make quick money, (he wants everything quick!) so he decided to rob banks. He wants to make a calculated risk, and grab as much money as possible. But his friends — Hermione and Ron have decided upon a tolerable probability P of getting caught. They feel that he is safe enough if the banks he robs together give a probability less than P.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case contains a real number P, the probability Harry needs to be below, and an integer N (0 < N ≤ 100), the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj (0 < Mj ≤ 100) and a real number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj. A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

Output

For each case, print the case number and the maximum number of millions he can expect to get while the probability of getting caught is less than P.

Sample Input

3

0.04 3

1 0.02

2 0.03

3 0.05

0.06 3

2 0.03

2 0.03

3 0.05

0.10 3

1 0.03

2 0.02

3 0.05

Sample Output:

Case 1: 2

Case 2: 4

Case 3: 6

Note:

For the first case, if he wants to rob bank 1 and 2, then the probability of getting caught is 0.02 + (1 — 0.02) * .03 = 0.0494 which is greater than the given probability (0.04). That's why he has only option, just to rob rank 2.

I had seen many solution in web where they calculated as:

for i = sum_of_millions to 1

    for j=1 to all bank

        dp[i][j] ....

My thought is why I should go for millions? Shouldn't I go by bank or not?

Tags dp
 
 
 
 
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5 years ago, # |
  Vote: I like it -10 Vote: I do not like it

Please don't help me by -ve votes.

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5 years ago, # |
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When harry potter rob two banks. The prob of getting caught in one of the two banks = the prob of getting caught in the first or the prob of not getting caught in the first and getting caught in the second.

Why we are going for millions instead of banks ? You can't build the dp solution by going into the banks instead of going for millions. If you wrote it down on paper how the dp solution is being built you'll notice that you'll always have a prob less than one just for the sum of millions that can be added from the banks. So for example in the 2nd sample test the prob of getting caught robbing 6 millions will be 1 even after building the dp solution because you cant really get to this state.

I hope I explained it in a simple way.

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5 years ago, # |
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Is Harry allowed to use Expecto Patronum?

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5 years ago, # |
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for i = 1 to all banks

for j = 1 to sum_of_ millions

dp[i][j]=...

I think it can work too.

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5 years ago, # |
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#include <bits/stdc++.h>
#define REP(i,a,b) for(int i=(a);i<=(b);i++)

using namespace std;

double Chance[102], Minimo[102][10002], Voldemort;
int Bancos, Casos, Plata[102], T;

int main()
{
    cin >> Casos;
    REP(caso,1,Casos)
    {
        cin >> Voldemort;
        cin >> Bancos;
        REP(i,1,Bancos) cin >> Plata[i] >> Chance[i];
        REP(i,0,Bancos) REP(j,0,10000) Minimo[i][j] = 1;
        int respuesta = 0;
        Minimo[0][0] = 0;
        REP(i,0,Bancos) REP(j,0,10000)
        {
            if(Minimo[i][j] <= Voldemort) respuesta = max(respuesta, j);
            if(i == Bancos) continue;

            // Robo el banco i+1

            int plata = j + Plata[i+1];
            Minimo[i+1][plata] = min(Minimo[i+1][plata], Minimo[i][j] + (1.0 - Minimo[i][j]) * Chance[i+1]);

            // No lo robo

            Minimo[i+1][j] = min(Minimo[i+1][j], Minimo[i][j]);
        }
        cout << "Case " << caso << ": " << respuesta << "\n";
    }
}