【PAT】 1085. Perfect Sequence (25)

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

分析：（1）int*int有可能会超出int范围，所以用long long （2）使用二分查找法，否则会超时

代码如下：

#include <iostream>#include <vector>#include <algorithm>using namespace std;//vec 采用“引用”则不会超时，如果用传参则会超时。 int binaryFind(vector<long long >& vec, long long max, int i){int left = i;int right = vec.size()-1;    int mid;    while(left < right){mid = (left+right)/2;if(vec[mid] > max){right = mid - 1;}else if(vec[mid] < max){left = mid + 1;}else{return mid;}}return vec[left]>max? left-1:left; }int main(int argc, char** argv) {long long  n, p, i, val;scanf("%lld%lld",&n,&p);vector<long long> vec(n);for(i=0; i<n; i++){scanf("%lld",&vec[i]);}sort(vec.begin(), vec.end());    long long max;    int cnt=0, pos;for(i=0; i<n; i++){max = p*vec[i];pos = binaryFind(vec, max, i); if(pos-i+1 > cnt){ cnt = pos-i+1;}}printf("%d\n",cnt);return 0;}