1085. Perfect Sequence (25)

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最新推荐文章于 2022-04-16 15:07:05 发布

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本文探讨了在给定整数序列和参数p的情况下，如何通过二分查找法找到序列中形成完美子序列的最大数量。重点在于理解完美序列的定义，并实现有效的算法来解决这一问题。

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Given a sequence of positive integers and another positive integer p. The sequence is said to be a “perfect sequence” if M <= m * 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 (<= 105) is the number of integers in the sequence, and p (<= 109) is the parameter. In the second line there are N positive integers, each is no greater than 109.

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:

1、考察二分法
2、注意溢出（可以标志负数最大，也可以使用long long 解决）

#include<iostream>
#include<algorithm>
using namespace std;
int a[1000005],n,p;
int find(int l,int t){
    int r=n-1;
    if(t<0)//溢出
        return r;
    while(r!=l){
        int mid=(r+l)/2;
        if(a[mid]>=t)
            r=mid;
        else l=mid+1;
    }
    while(a[r]>t)
        r--;
    return r;
}
int main(){
    cin>>n>>p;
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n);
    int maxl=0;
    for(int i=0;i<n;i++)
    {
        int M=a[i]*p;
        int k=find(i,M);
        maxl=max(k-i+1,maxl);
    }
    cout<<maxl<<endl;
    return 0;
}

使用long long解决

#include<iostream>
#include<algorithm>
using namespace std;
long long a[1000005],n,p;
int find(long long l,long long t){
    int r=n-1;
    while(r!=l){
        int mid=(r+l)/2;
        if(a[mid]>=t)
            r=mid;
        else l=mid+1;
    }
    while(a[r]>t)
        r--;
    return r;
}
int main(){
    freopen("in.txt","r",stdin);
    cin>>n>>p;
    for(int i=0;i<n;i++)
        cin>>a[i];
    sort(a,a+n);
    int maxl=0;
    for(int i=0;i<n;i++)
    {
        long long M=a[i]*p;
        int k=find(i,M);
        maxl=max(k-i+1,maxl);
    }
    cout<<maxl<<endl;
    return 0;
}