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 9Sample Output:
8
先排个升序,然后用二分,对前一半的数找出最大的M,更新最大数m。
代码:
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { long long s[100000],n,p,m = 0; scanf("%lld%lld",&n,&p); for(int i = 0;i < n;i ++) { scanf("%lld",&s[i]); } sort(s,s + n); for(int i = 0;i < n/2;i ++) { int l = i,r = n - 1,mid = (l + r)/2; while(l < r) { if(s[mid] > s[i] * p)r = mid - 1; else if(s[mid] < s[i] * p)l = mid + 1; else break; mid = (l + r) / 2; } if(s[mid] > s[i] * p)mid --; if(m < mid - i)m = mid - i; } printf("%lld",m + 1); }