问题
1085 Perfect Sequence (25 分)
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 (≤10
5
) is the number of integers in the sequence, and p (≤10
9
) is the parameter. In the second line there are N positive integers, each is no greater than 10
9
.
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
结尾无空行
Sample Output:
8
思路
对数组从小到大遍历，然后枚举每一个元素，使之与p相乘，利用upper_bound找到第一个大于该乘积的元素。更新元素个数的最大值。
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int seq[maxn];
int main(){
int n,p,temp,ans=1;
cin>>n>>p;
for (int i = 0; i < n; ++i) {
cin>>temp;
seq[i] = temp;
}
sort(seq,seq+n);
for (int j = 0; j < n; ++j) {
int x = upper_bound(seq+j+1,seq+n,(long long)seq[j]*p)-seq;
ans = max(ans,x-j);
}
cout<<ans;
return 0;
}
总结
记住upper_bound()函数和lower_bound()函数的用法，可以简化大量的代码。