标签：advance pat
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 9Sample Output:
8
#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);
}
Pat(Advanced Level)Practice--1085(Perfect Sequence)
标签：advance pat
原文地址：http://blog.csdn.net/cstopcoder/article/details/42566243
