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?yàn)閍[i]和p都可能達(dá)到10^9,所以mp可達(dá)到10^18,因此用long long,用int則有一個(gè)測(cè)試點(diǎn)錯(cuò)誤 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結(jié)束時(shí)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是運(yùn)用二分原理，查找a數(shù)組中第一個(gè)大于a[i]*p的元素的指針 //因此這里也可以自己寫一個(gè)binarySearch(int i,long long x)函數(shù) int j=binarySearch(i,n,(long long)a[i]*p); ans=max(ans,j-i); } printf("%d/n",ans); return 0;}新聞熱點(diǎn)
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