PAT A1085

最新推荐文章于 2020-06-10 15:41:21 发布

xiao1guaishou

最新推荐文章于 2020-06-10 15:41:21 发布

阅读量199

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT 二分法

本文链接：https://blog.youkuaiyun.com/xiao1guaishou/article/details/89053057

PAT 同时被 2 个专栏收录

60 篇文章

订阅专栏

二分法

5 篇文章

订阅专栏

这是一道编程题，要求找出一个序列中满足最大值不超过最小值p倍的最长完美子序列。给定序列长度N和参数p，以及序列本身，通过排序和双指针或二分法求解。样例输入为10 8和2 3 20 4 5 1 6 7 8 9，输出结果为8。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：
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:

题目大意
从n个正整数中选择若干个数，使得选出的这些数中最大值不超过最小值的p倍。问满足条件的选择方案中，选出的数的最大个数
方法一：双指针法
分析：
1.对输入的数组排序
2.对每个1~n-1个数，i指向开头的数， j指向尾部的数，若j < n && A[j] <= (long long)A[i]*p，记录下j-i+1，
3.输出j-i+1中最大值

代码实现：

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int Max = 100010;
int main()
{
    int A[Max];
    int n, p;
    cin >> n >> p;
    for(int i = 0; i < n; i++)
        scanf("%d", &A[i]);
    sort(A, A+n);                 //对A数组从小到大排序
    int ans = 0;
    int i = 0, j = 0;
    while(i < n && j< n){
        while(j < n && A[j] <= (long long)A[i]*p)
        {
            j++;
            ans = max(ans, j-i+1);
        }
        i++;
    }
    cout << ans;
    return 0;


}

方法二：二分法
分析：对于每个A[i],只用找到第一个A[j],使得A[j] > A[i]*p;输出最大的j-i;

代码实现：

#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
int A[100010], n, p;
int binSearch(int i, long long x ){
    if(A[n-1] < x)
        return n;
    int l = i+1, r = n-1, mid;
    while(l < r){
        mid = (l+r)/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);                     //对数组A从小到大排序
    int ans = 0;
    for(int i =0; i < n; i++){
        int j = binSearch(i, (long long)A[i] * p);
        ans = max(ans, j-i);
    }
    printf("%d", ans);
    return 0;
}