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Pat(Advanced Level)Practice--1085(Perfect Sequence)
Pat1085代码
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); }
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???
Pat(Advanced Level)Practice--1085(Perfect Sequence)