Pat(Advanced Level)Practice--1085(Perfect Sequence)

Pat1085代码

题目描述：

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

AC代码：

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#define MAXN 100005

using namespace std;

double v[MAXN];

int main(int argc,char *argv[]){
	int n;
	double p;
	int i,j;
	scanf("%d%lf",&n,&p);
	for(int i=0;i<n;i++)
		scanf("%lf",&v[i]);
	sort(v,v+n);
	int maxlen=0;
	for(i=0;i<n;i++){
		for(j=i+maxlen-1;j<n;j++){
			double temp=v[i]*p;
			if(v[j]>temp)
				break;
			if(maxlen<j-i+1)
				maxlen=j-i+1;
		}
	}
	printf("%d\n",maxlen);
}

疑惑：题目明明说的n,p和数组元素都是正整数，为什么只有换成double之后最后一个case才能通过，难道测试数据和题目描述不符？？？