1085 Perfect Sequence

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 (≤) 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.

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

题意：

给出一组数，在这组数中尽可能多的选出数字，使得所选出数字的最大值M和最小值n，满足M <= n * p。刚开始的时候我就理解错了，没有搞明白题意，以为是按照所给出序列的顺序来寻找。

思路：

https://www.liuchuo.net/archives/1908

Code:

#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int main() {
    int t;
    long long p;
    cin >> t >> p;
    int c;
    vector<int> v;
    for (int i = 0; i < t; ++i) {
        cin >> c;
        v.push_back(c);
    }
    sort(v.begin(), v.end());
    int temp = 0, result = 0;
    for (int i = 0; i < t; ++i) {
        for (int j = i+result; j < t; ++j) {
            if (v[j] <= v[i]*p) {
                temp = j - i + 1;
                if (temp > result) 
                    result = temp;
            } else {
                break;
            }
        }
    }
    cout << result << endl;
    return 0;
}

　　

int的取值范围：-2147483648 到2147483647

这里p的范围是小于10⁹已经超出了int的范围，所以应该选择long long

            if (v[j] <= v[i]*p) {
                temp = j - i + 1;
                if (temp > result) 
                    result = temp;
            }        

　　v[i] * p 会不会溢出成为负值呢？