标签：otto include arc sed printf ext html math long
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.
Each input file contains one test case. For each case, the first line contains two positive integers N and p, where N (≤) is the number of integers in the sequence, and p (≤) is the parameter. In the second line there are N positive integers, each is no greater than 1.
For each test case, print in one line the maximum number of integers that can be chosen to form a perfect subsequence.
10 8
2 3 20 4 5 1 6 7 8 9
8
#include<iostream> #include<algorithm> using namespace std; int binarysearch(int a[], long long x,int size,int i) {//查找第一个大于x的数的位置 int left = i+1, right = size-1, mid; if(a[size-1]<=x) return size; while(left < right) { mid = (left + right)/2; if(a[mid]<=x) left = mid+1; else right = mid; } return left; } int main() { long long n,p; scanf("%lld %lld",&n,&p); int a[n]; for(int i=0;i<n;i++) { scanf("%d",&a[i]); } sort(a,a+n); int cnt =0; for(int i=0;i<n;i++) {
//i从0——n-1，寻找大于p*a[i]的数在数组中的下标 int j = binarysearch(a,a[i]*p,n,i)-i;//a[i]是最小值 if(cnt<j) cnt = j; } cout<<cnt; return 0; }
解法二：two_point大法
#include<iostream> #include<algorithm> using namespace std; int main() { int n,p; scanf("%d %d",&n,&p); long long a[n]; for(int i=0;i<n;i++) { cin>>a[i]; } sort(a,a+n); int cnt =1; int i=0,j=0;//i，j是两个下标或者说是指针，i，j分别向右边移动 while(i<n&&j<n) { while(j<n&&a[j]<=a[i]*p)//满足条件j就往后移动 { cnt=max(cnt,j-i+1);//cnt取最大值，所以j不需要从i+1开始 j++; } i++; } printf("%d",cnt); return 0; }
标签：otto include arc sed printf ext html math long
原文地址：https://www.cnblogs.com/qinmin/p/12771174.html
