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.
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.
For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.
思路： 首先将数字序列按照非递减的顺序排序，从第一个数开始计算完美数列的长度，最后取最大值。在计算完美数列的时候，要用二分查找的策略，找到第一个大于x*p的位置，将时间复杂度由n变为logn。
代码一：手写二分查找的函数
#include <iostream> #include<algorithm> using namespace std; const int maxn = 100010; int a[maxn],n,p; int BinarySearch(int i,long long x) { if (a[n - 1] <= x) return n; int l = i + 1, r = n - 1, mid; while (l<r) { mid = (l + r) / 2; if (a[mid] <= x) l = mid + 1; else { r = mid; } } return l; } int main() { 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 j = BinarySearch(i, (long long)a[i] * p); ans = max(ans, j - i); } printf("%d\n", ans); return 0; }
代码二：利用STL模板库的upper_bound函数
#include <iostream> #include<algorithm> using namespace std; const int maxn = 100010; int a[maxn],n,p; int main() { 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 j= upper_bound(a + i+1, a + n,(long long) a[i] * p)-a; ans = max(ans, j - i); } printf("%d\n", ans); return 0; }
