Robberies

The aspiring Roy the Robber has seen a lot of American movies, and knows that the bad guys usually gets caught in the end, often because they become too greedy. He has decided to work in the lucrative business of bank robbery only for a short while, before retiring to a comfortable job at a university.

For a few months now, Roy has been assessing the security of various banks and the amount of cash they hold. He wants to make a calculated risk, and grab as much money as possible. His mother, Ola, has decided upon a tolerable probability of getting caught. She feels that he is safe enough if the banks he robs together give a probability less than this.

The first line of input gives T, the number of cases. For each scenario, the first line of input gives a floating point number P, the probability Roy needs to be below, and an integer N, the number of banks he has plans for. Then follow N lines, where line j gives an integer Mj and a floating point number Pj . Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .

For each test case, output a line with the maximum number of millions he can expect to get while the probability of getting caught is less than the limit set. Notes and Constraints 0 < T <= 100 0.0 <= P <= 1.0 0 < N <= 100 0 < Mj <= 100 0.0 <= Pj <= 1.0 A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.

30.04 31 0.022 0.033 0.050.06 32 0.032 0.033 0.050.10 31 0.032 0.023 0.05

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/*转载！     转成以所有银行的总资产为背包容量V。。求最大的逃跑概率。。注意：题目给我们的是被抓的概率，，而我们要求最大的逃跑率，需要去被抓的概率pi的补 ，即1-pi只有逃跑率才会等于各个逃跑率之积，被抓的概率不会等于各个被抓的概率之积，，概率的知识，不多说。。状态转移方程：dp[j] = max ( dp[j], dp[j-cost[i]] * weight[i])..*/ #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;const int N=100*100+10;double dp[N]; //  拿了i钱的逃跑概率dp[i] int sum;int cost[100+5];double p[100+5];int n;double f;int main(){     int t,i,j;    scanf("%d",&t);    while(t--){        sum=0;        scanf("%lf%d",&f,&n);        for(i=0;i
         
          =cost[i];j--) dp[j]=max(dp[j],dp[j-cost[i]]*(1.0-p[i])); /* 注释这里是错的 01背包必须从后往前更新 因为后面会涉及到前面的 而01背包i状态时根据i-1状态推过来的 如果从前推 前面的状态就会变成i而不是i-1 到后面时就找不到i-1的状态了 for(j=0;j<=sum;j++){ if(j>=cost[i]) else continue; } */ for(i=sum;i>=0;i--){ if(dp[i]>1.0-f){ printf("%d\n",i); break; } } } return 0;}