1085. Perfect Sequence (25)二分查找——PAT (Advanced Level) Practise

Posted 2020-06-25 闲云阁

题目信息

1085. Perfect Sequence (25)

时间限制300 ms
内存限制65536 kB
代码长度限制16000 B
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 (<= 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

解题思路

排序后二分查找即可

AC代码

#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\n", cnt);
    return 0;
}