1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

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提交代码

题目意思：

给定一个正整数数列，和正整数p，设这个数列中的最大值是M，最小值是m，如果M <= m * p，则称这个数列是完美数列。

现在给定参数p和一些正整数，请你从中选择尽可能多的数构成一个完美数列。

输入格式：

输入第一行给出两个正整数N和p，其中N（<= 10⁵）是输入的正整数的个数，p（<= 10⁹）是给定的参数。第二行给出N个正整数，每个数不超过10⁹。

输出格式：

在一行中输出最多可以选择多少个数可以用它们组成一个完美数列。

注意：M不一定是里面最大的数，可以是任意数，题目是要我们选出一些数，而不是所有数

#include<cstdio>#include<algorithm>using namespace std;#define N 100005int main(){int n,p,a[N];//freopen("input.txt","r",stdin);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 index = upper_bound(a+i+1,a+n,(long long)a[i]*p) - a;ans = max(ans,index - i);if(ans > n-i){break;}}printf("%d\n",ans);return 0;}