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Problem Description
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.<br><br><center><img src=../../../data/images/con211-1010-1.jpg></center> <br>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.<br><br><br>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.
 

Input
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 . <br>Bank j contains Mj millions, and the probability of getting caught from robbing it is Pj .
 

Output
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.<br><br>Notes and Constraints<br> 0 < T <= 100<br> 0.0 <= P <= 1.0<br> 0 < N <= 100<br> 0 < Mj <= 100<br> 0.0 <= Pj <= 1.0<br> A bank goes bankrupt if it is robbed, and you may assume that all probabilities are independent as the police have very low funds.
 

Sample Input
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
 

Sample Output
24

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题目大意:求一个奇葩小偷能偷到的最大的钱给你他被抓的概率和最大的被抓概率

思路:0 1背包,但是问题是没法存的是答案。。概率问题。被抓概率是p那不被抓就是1-pso要都不被抓药用乘法原理是(1-p1)*(1-p2)这个概率是不被抓那被抓的是1-刚刚概率 所以被抓小于p 所以被抓的要小于刚刚的概率。这就比较奇葩了需要存概率。。于是我就换了个想法把概率当做价值求出所有的概率最后比较概率满足条件的输出钱

还有个奇葩地方。。写完后超时,我把所有c++输入输出换成c还是超时。。后来百度下发现这个题大多都是running time 因为钱很多所以不能用105然后我该了数组长度后

超时解决了。。真心搞不懂oj的检查机制太奇葩了。

代码:#include<iostream>#include<stdio.h>#include<string.h>using namespace std;struct node {  int money;   double runp; }bank[105];  double dp[10005];int main(){    int t;     scanf("%d",&t);    while (t--)    { int   maxmoney=0;        double p;        int n;        scanf("%lf %d",&p,&n);         p=1-p;        for(int i=0;i<n;i++)        {        scanf("%d %lf",&bank[i].money,&bank[i].runp);        bank[i].runp=1-bank[i].runp;        maxmoney+=bank[i].money;        }      memset(dp,0,sizeof(dp));    dp[0]=1.0;            for(int i=0;i<n;i++){        for(int j=maxmoney;j>=bank[i].money;j--)        dp[j]=max(dp[j-bank[i].money]*bank[i].runp,dp[j]);        }        for(int i=maxmoney;i>=0;i--)        {                       if(dp[i]>p)                       {                            printf("%d\n",i);                           break;                           }            }}    return 0;    }

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