hdu 2955 Robberies

Robberies

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5253    Accepted Submission(s): 1993

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.


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

 

Source

IDI Open 2009

 

Recommend

gaojie

觉得这个题的主要难点在于理解题意，与一般的01背包问题不同，还需进行转换

讲被抓的概率转换为不被抓的概率。只有在不被抓的情况下才进行下一场抢劫！

然后是小数，感觉上没有整数好处理，应为存在浮点误差，幸好这个题没为难浮点误差.

#include<stdio.h>#include<math.h>#define MAXN 101 int c[MAXN];double p[MAXN],f[10001];int main(void){int T;//freopen("d:\\in.txt","r",stdin);scanf("%d",&T);while(T--){double P;int N,i,j;scanf("%lf%d",&P,&N);P=1-P;int s=0;for(i=1;i<=N;i++){scanf("%d%lf",&c[i],&p[i]);p[i]=1-p[i];s+=c[i];}for(i=s;i>=1;i--)f[i]=0;f[0]=1;for(i=1;i<=N;i++)for(j=s;j>=c[i];j--){if(f[j]<f[j-c[i]]*p[i] &&f[j-c[i]]*p[i]>=P)f[j]=f[j-c[i]]*p[i];}for(i=s;i>=0;i--)if(f[i]>=P)break;printf("%d\n",i);}return 0;}