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 8
2 3 20 4 5 1 6 7 8 9
Sample Output:
8

//法1：#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;int a[maxn];bool cmp(int a,int b){    return a<b;}int main(){    int n,p;    scanf("%d%d",&n,&p);    for(int i=0;i<n;i++){        scanf("%d",&a[i]);    }    sort(a,a+n,cmp);    int k=0,maxinum=0;    for(int i=0;i<n;i++){        long long mp=(long long)a[i]*p;//因为a[i]和p都可能达到10^9,所以mp可达到10^18,因此用long long,用int则有一个测试点错误         while(a[k]<=mp&&k<n){            k++;        }        maxinum=max(maxinum,k-i);    }    printf("%d\n",maxinum);}//法2：#include<cstdio>#include<algorithm>using namespace std;const int maxn=100010;int a[maxn];int binarySearch(int i,int n,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;//由于while结束时l==r,因此返回l或者r都行 }int main(){    int n,p;    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;//upper_bound是运用二分原理，查找a数组中第一个大于a[i]*p的元素的指针        //因此这里也可以自己写一个binarySearch(int i,long long x)函数        int j=binarySearch(i,n,(long long)a[i]*p);         ans=max(ans,j-i);    }     printf("%d\n",ans);    return 0;}