pat 1085 Perfect Sequence

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

#include <iostream>#include <cstdio>#include <vector>#include <algorithm>using namespace std;vector<long long> vec;long long binarySearch(long long istart, long long iend, long long num){while(istart <= iend) {long long mid = (istart + iend) / 2;if(vec[mid] == num) return mid;else if(vec[mid] > num) iend = mid-1;else istart = mid+1;}return istart;}int main() {long long n, p;scanf("%lld %lld", &n, &p);for (long long i = 0; i < n; ++i){long long x;scanf("%lld", &x);vec.push_back(x);}sort(vec.begin(), vec.end());long long ans = 0;for (long long i = 0; i < n; ++i){long long x = p * vec[i];long long pos = binarySearch(i+1, vec.size()-1, x);long long tmp = pos - i;if(pos < n && vec[pos] == x) tmp += 1; //如果x在数组内,加x加入if(tmp > ans) ans = tmp;}printf("%lld\n", ans);return 0;}