pat 1085. Perfect Sequence (25)

pat 1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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 9

Sample Output:

8

这道题目，在时间上，就考察一个二分查找。还有就是一个long long因为乘法有可能超过int

#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
vector<long long> nums;
int bSearch(long long num,int n)
{
    int l = 0,r = n-1,mid;
    while(l<=r)
    {
        mid = (l+r)/2;
        if(nums[mid]>num)
        {
            r = mid-1;
        }else if(nums[mid]<num)
        {
            l = mid+1;
        }else
        {
            return mid;
        }
    }
    return l;
}
int main()
{
    long long n,p,tmp1,m,index;
    long long i,j,tmpMax=0,resMax;
    scanf("%lld%lld",&n,&p);
    for(i=0;i<n;i++)
    {
        scanf("%lld",&tmp1);
        nums.push_back(tmp1);
    }
    sort(nums.begin(),nums.end());
    for(i=0;i<n;i++)
    {
        m = nums[i]*p;
        index = bSearch(m, n);
        if(nums[n-1]<=m)
        {
            tmpMax = n-1-i+1;
        }else
        {
            tmpMax = index-i;
        }
        if(tmpMax>resMax)
        {
            resMax = tmpMax;
        }
    }
    printf("%lld\n",resMax);
    return 0;
}