pat1085. Perfect Sequence (25)

1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 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

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提交代码

#include<cstdio>
#include<stack>
#include<algorithm>
#include<iostream>
#include<stack>
#include<set>
#include<map>
#include<vector>
using namespace std;
long long line[100005];
int main(){
    //freopen("D:\INPUT.txt","r",stdin);
    long long n,p;
    scanf("%lld %lld",&n,&p);
    int i,j;
    for(i=0;i<n;i++){
        scanf("%lld",&line[i]);
    }
    sort(line,line+n);
    int amount;
    long long cur=line[0]*p;
    for(j=0;j<n&&line[j]<=cur;j++);//j指针
    amount=j;
    for(i=1;i<n&&j<n;i++){//这里的j<n是为了避免不必要的循环
        cur=line[i]*p;
        for(;j<n&&line[j]<=cur;j++);
        if(j-i>amount){
            amount=j-i;
        }
    }
    printf("%d
",amount);
    return 0;
}