1085. Perfect Sequence(25)[2점 찾기] - PAT(Advanced Level) Practise

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1085. Perfect Sequence (25)

시간 제한 300ms 메모리 제한 65536kB 코드 길이 제한 16000B 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 minimumbers 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 (<= 10^5) is the number of integers in the sequence, and p (<= 10^9) is the parameter. In the second line there are N positive integers, each is no greater than 10^9.
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

문제풀이의 방향

정렬 후 2분 검색하시면 됩니다.

AC 코드

#include <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int main(){
    long long n, p, t;
    vector<long long> v;
    scanf("%lld%lld", &n, &p);
    for (long long i = 0; i < n; ++i){
        scanf("%lld", &t);
        v.push_back(t);
    }
    sort(v.begin(), v.end());
    int cnt = 0;
    for (int i = 0; i < v.size(); ++i){
        t = p * v[i];
        int tmp = distance(v.begin(), upper_bound(v.begin(), v.end(), t)) - i;
        cnt = max(cnt, tmp);
    }
    printf("%d
", cnt);
    return 0;
}