PAT Advanced 1085 Perfect Sequence (25) [⼆分,two pointers]
题目
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
题目分析
已知正整数序列seq[N],最大值为M,最小值为m,已知另一个正整数p(<=10^9),从数列中抽出一部分数字,求可以满足M<=m*p的数字最多抽取个数
要满足M<=mp抽取的数字最多(即:M与m中间夹的数字最多),需要取所有满足M<=mp的情况中,m最小,M最大
解题思路
思路 01(最优、二分查找、查找M复杂度O(logn))
- 对seq[N]升序排序
- 依次遍历seq[i],在i+1到N之间,找到最大满足M<=mp的数字(即:第一个满足大于mp的数字下标j-1)
思路 02 (two pointer、查找M复杂度O(n))
- 对seq[N]升序排序
- 依次遍历seq[i],j初始为0,开始从上次j往后找(因为i+1后m增大,m*q>=M,所以M增大,j只能在上次j之后)
易错点
- p(<=10^9),所以m*p有可能超过int范围,数组元素类型需为long long,否则第5个测试点错误
- 取第一个大于mp的数字下标-1,而不是第一个大于等于mp的数字下标(因为大于的情况下要-1,等于的情况下不需要-1,处理麻烦)
- 思路02中,只能从前往后找第一个不满足条件m*q>=M的,不能从后往前找最后一个满足条件的(测试点4超时)
Code
Code 01
#include <iostream>
#include <algorithm>
using namespace std;
int main(int argc,char * argv[]) {
int n,p;
scanf("%d %d",&n,&p);
long long seq[n]= {0}; // 若为int,第5个测试点错误
for(int i=0; i<n; i++) {
scanf("%d",&seq[i]);
}
sort(seq,seq+n);
int maxnum=0;
for(int i=0; i<n; i++) {
// 二分查找
int left=i+1,right=n;
int mid = left+((right-left)>>1);
while(left<right) {
mid = left+((right-left)>>1);
if(seq[mid]>seq[i]*p) { //若是求第一个大于等于seq[i]*p,测试点2错误
right=mid;
} else {
left=mid+1;
}
}
if(right-i>maxnum)maxnum=right-i;
}
printf("%d",maxnum);
return 0;
}
Code 01
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
int main(int argc,char * argv[]) {
int n,p;
scanf("%d %d",&n,&p);
long long seq[n]= {0}; // 若为int,第5个测试点错误
for(int i=0; i<n; i++) {
scanf("%d",&seq[i]);
}
sort(seq,seq+n);
// 写法一:
int maxnum=0,j = 0;
for(int i=0; i<n; i++) {
while(j<n&&seq[i]*p>=seq[j]) j++;
maxnum=max(maxnum,j-i);
}
// 写法二:
// int i=0,j=0,maxnum=1;
// while(i<n&&j<n) {
// while(j<n&&seq[j]<=(long long)seq[i]*p) {
// maxnum=max(maxnum,j-i+1);
// j++;
// }
// i++;
// }
/*
使用下面代码,第四个测试点超时
j从后往前找最后一个满足条件的,测试点4超时
*/
// int maxnum=0,prej=0; //prej用于记录上次j的位置,之后的j只可能比prej大,m*p>=M;i+1因为m增大了,所以M一定增大
// for(int i=0; i<n; i++) {
// int j = n-1;
// while(prej<=j&&seq[i]*p<seq[j]) j--;
// maxnum=max(maxnum,j-i+1);
// prej=j;
// }
printf("%d",maxnum);
return 0;
}