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  • 1085 Perfect Sequence (25 分)

    Given a sequence of positive integers and another positive integer p. The sequence is said to be a perfect sequence if Mm×p where M and m are the maximum and minimum numbers in the sequence, respectively.

    Now given a sequence and a parameter p, you are supposed to find from the sequence as many numbers as possible to form a perfect subsequence.

    Input Specification:

    Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.

    Output Specification:

    For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.

    Sample Input:

    10 8
    2 3 20 4 5 1 6 7 8 9
     

    Sample Output:

    8
     
     
    #include<bits/stdc++.h>
    using namespace std;
    const int maxn=100010;
    
    int n,p;
    int a[maxn];
    int binarySearch(int i,long long x){
        if(a[n-1]<=x){
            return n;
        }
        int L=i+1,r=n-1,mid;
        while(L<r){
            mid=(r+L)/2;
            if(a[mid]<=x){
                L=mid+1;
            }else{
                r=mid;
            }
        }
        return L;
    }
    int main(){
        scanf("%d %d",&n,&p);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        sort(a,a+n);
        int ans=1;
        for(int i=0;i<n;i++){
            int j=binarySearch(i,(long long)a[i]*p);
            ans=max(ans,j-i);
        }
        printf("%d
    ",ans);
        return 0;
    }

    扛不住了,要睡觉。。。

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  • 原文地址:https://www.cnblogs.com/dreamzj/p/14472112.html
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