HDU 4815 2013長春現場賽C題

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C - Little Tiger vs. Deep Monkey Time Limit:1000MS      Memory Limit:65535KB      64bit IO Format:%I64d & %I64uSubmit Status Practice HDU 4815

Description

A crowd of little animals is visiting a mysterious laboratory ? The Deep Lab of SYSU. 

“Are you surprised by the STS (speech to speech) technology of Microsoft Research and the cat face recognition project of Google and academia? Are you curious about what technology is behind those fantastic demos?” asks the director of the Deep Lab. “Deep learning, deep learning!” Little Tiger raises his hand briskly. “Yes, clever boy, that’s deep learning (深度學習/深度神經網路)”, says the director. “However, they are only ‘a piece of cake’. I won’t tell you a top secret that our lab has invented a Deep Monkey (深猴) with more advanced technology. And that guy is as smart as human!” 

“Nani ?!” Little Tiger doubts about that as he is the smartest kid in his kindergarten; even so, he is not as smart as human, “how can a monkey be smarter than me? I will challenge him.” 

To verify their research achievement, the researchers of the Deep Lab are going to host an intelligence test for Little Tiger and Deep Monkey. 

The test is composed of N binary choice questions. And different questions may have different scores according to their difficulties. One can get the corresponding score for a question if he chooses the correct answer; otherwise, he gets nothing. The overall score is counted as the sum of scores one gets from each question. The one with a larger overall score wins; tie happens when they get the same score. 

Little Tiger assumes that Deep Monkey will choose the answer randomly as he doesn’t believe the monkey is smart. Now, Little Tiger is wondering “what score should I get at least so that I will not lose in the contest with probability of at least P? ”. As little tiger is a really smart guy, he can evaluate the answer quickly. 

You, Deep Monkey, can you work it out? Show your power!?/div> 

Input

The first line of input contains a single integer T (1 ≤ T ≤ 10) indicating the number of test cases. Then T test cases follow. 

Each test case is composed of two lines. The first line has two numbers N and P separated by a blank. N is an integer, satisfying 1 ≤ N ≤ 40. P is a floating number with at most 3 digits after the decimal point, and is in the range of [0, 1]. The second line has N numbers separated by blanks, which are the scores of each question. The score of each questions is an integer and in the range of [1, 1000]?/div> 

Output

For each test case, output only a single line with the answer. 

Sample Input

 13 0.51 2 3 

 

Sample Output

 3 

這題逗逼了，剛開始大帝告訴我題意可能我理解錯了，然後就用二進位枚舉了！然後然後……然後就浪費了一個多小時一直WA，後面大帝發覺我理解錯了之後，他又敲了背包才過……唉……發現題意真的是有點難理解了。到現在題意和解法還都是半知半懂的。

正確的解法是：總的情況是:1<<n，然後可組合的次數除以總的情況>=p的最小分數是正確答案。

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;long long dp[40005];int main(){    int t;    cin>>t;    while(t--)    {        int n,i,j,sum=0,a[45];        double p;        memset(dp,0,sizeof(dp));        dp[0]=1;        cin>>n>>p;        for(i=0;i<n;i++)            scanf("%d",a+i),sum+=a[i];        sort(a,a+n);        for(i=0;i<n;i++)            for(j=sum;j>=a[i];j--)                dp[j]+=dp[j-a[i]];        long long sum1=1LL<<n,sum2=0;        for(i=0;i<=sum;i++)        {            sum2+=dp[i];            if((double)sum2/(double)sum1>=p)            {                printf("%d\n",i);                break;            }        }    }    return 0;}