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
总结
二分 #two_pointers 可以用二分、two pointers两种方法。
- 二分有两种做法:
- 二分答案;
- 对排序后的每一个
a[i]
,在a[i+1]~a[n-1]
内查找第一个超过a[i]*p
的数的位置j。
- two pointers:i=j=0,让j在符合条件的情况下不断增大,不能增大时则令i++,时间复杂度只需 $o(n)$ 。
// 二分答案
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,p, e[100000+5];
int main(){
scanf("%lld%lld", &n, &p);
for(int i = 0; i < n; i++) scanf("%lld", &e[i]);
sort(e, e+n);
int lo = 0, hi = n, mi=0;
while(hi > lo){
mi = (hi + lo)>>1;
int en = 0;
for(int i = 0; i + mi < n; i++){
if(e[i + mi] <= e[i]*p){
en = 1;
break;
}
}
if(en) lo = mi+1;
else hi = mi;
}
printf("%d", mi+1);
return 0;
}
//二分数组
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,p, e[100000+5];
int main(){
scanf("%lld%lld", &n, &p);
for(int i = 0; i < n; i++) scanf("%lld", &e[i]);
sort(e, e+n);
int r = 1;
for(int i = 0; i < n; i++){
int j = upper_bound(e+i+1, e+n, (ll)e[i]*p) - e;
r = max(r, j-i);
}
printf("%d", r);
return 0;
}
//two pointers
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,p, e[100000+5];
int main(){
scanf("%lld%lld", &n, &p);
for(int i = 0; i < n; i++) scanf("%lld", &e[i]);
sort(e, e+n);
int r = 0;
int i = 0, j = 0;
while(i < n && j < n){
while(j < n && e[j] <= e[i]*p){
r=max(r, j-i+1);
j++;
}
i++;
}
printf("%d", r);
return 0;
}
标签:Perfect,1085,PAT,sequence,int,ll,long,++,scanf
From: https://www.cnblogs.com/Afinis/p/17776457.html