1030 perfect sequence (25 points)
Given a positive integer sequence and a positive integer p , let the maximum value in this sequence be M and the minimum value be m . If M ≤ mp , then this sequence is called a perfect sequence.
Now given the parameter p and some positive integers, please select as many numbers as possible to form a perfect sequence.
Input format:
The first line of input gives two positive integers N and p , where N (≤105
) is the number of positive integers entered, p (≤109
) is the given parameter. The second line gives N positive integers, each number does not exceed 109
。
Output format:
Output in one line the maximum number of numbers that can be selected to form a perfect sequence.
Input example:
10 8
2 3 20 4 5 1 6 7 8 9
Output sample:
8
Question analysis:
Note that p and the given N positive integers are both <10^9, and long storage should be used. (last test point)
① Sort N positive integers from small to large 1 2 3 4 5 6 7 8 9 20 nums[] array storage
② Start looping from left to right, find the subscript of j where num[j]>nums[i]*p, and then use ji to find out how many numbers there are in total. Attached code:
//看这段代码的时候可以拿出纸和笔来操作一下，是整道题的核心
int max=0,result=0;//max记录最大个数，result中间变量
for(int i=0;i<N;i++){
//设置i+max可以缩短循环时间，max比如为4，
// 只需考虑i+4之后的数，之前的数一定小于4位
for(int j=i+max;j<N;j++){
if(nums[j]<=nums[i]*p){
result=j-i+1;//记录一下当前j和i之间有几个数
//如果max小于result，令max=result
if(max<result)
max=result;
}else //这说明nums[j]>nums[i]*p,不满足M<=mp,结束循环
break;
}
}
③Output max value
without AC code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
public class Main {
public static void main(String[] args) throws Exception{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String[] str=br.readLine().split(" ");
int N=Integer.parseInt(str[0]);
//用long存储
long p=Long.parseLong(str[1]);
String[] st=br.readLine().split(" ");
long nums[]=new long[N];//用long
for(int i=0;i<N;i++){
nums[i]=Long.parseLong(st[i]);
}
Arrays.sort(nums);//排序
//这道题的核心代码
int max=0,result=0;
for(int i=0;i<N;i++){
for(int j=i+max;j<N;j++){
if(nums[j]<=nums[i]*p){
result=j-i+1;
if(max<result)
max=result;
}else
break;
}
}
//输出
System.out.print(max);
}
}
My method of finding max for this question is already very fast. I tried to switch to a faster sorting method, but test point 4 still times out. The unfriendliness of PAT towards JAVA is once again reflected. Again, if this happens to everyone, don’t waste time deducting full points.
If there is a faster way, please leave a message in the comment area.
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