hdu 2955

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.


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.

 

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 .
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.

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.

 

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

246 
题意：
输入一个数，表示一共有多少组数据，输入一个浮点型数据表示被小偷被抓的几率不超过这个的情况下能头的最多钱，输入一个整型表示银行数，求小偷在不被抓的情况下头的最多钱。
思路：
背包;第一次做的时候把概率当做背包(放大100000倍化为整数):在此范围内最多能抢多少钱  最脑残的是把总的概率以为是抢N家银行的概率之和… 把状态转移方程写成了f[j]=max{f[j],f[j-q[i].v]+q[i].money}(f[j]表示在概率j之下能抢的大洋);正确的方程是:f[j]=max(f[j],f[j-q[i].money]*q[i].v)  其中,f[j]表示抢j块大洋的最大的逃脱概率,条件是f[j-q[i].money]可达,也就是之前抢劫过; 始化为:f[0]=1,其余初始化为-1  (抢0块大洋肯定不被抓嘛)
代码：
#include<cstdio>#include<cstring>#include<algorithm>int m[1000];//银行里的钱double pp[1000];//抢劫银行被抓的概率double dp[10000];using namespace std;int main(){    int T,N;    double P;    scanf("%d",&T);    while(T--)    {        int sum=0;        scanf("%lf%d",&P,&N);        for(int i=0; i<N; i++)        {            scanf("%d%lf",&m[i],&pp[i]);            sum+=m[i];        }        memset(dp,0,sizeof(dp));        dp[0]=1;//抢劫银行0元钱不被抓的概率为1        for(int i=0; i<N; i++)            for(int j=sum; j>=m[i]; j--)                dp[j]=max(dp[j],dp[j-m[i]]*(1-pp[i]));                          for(int i=sum; i>=0; i--)            {                if(dp[i]>=1-P)                {                    printf("%d\n",i);                    break;                }            }    }    return 0;}