1085. Perfect Sequence (25)

时间限制

300 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<algorithm>using namespace std;const int maxn=1000007;typedef long long ll;template<class T>inline T read(T&x){    char c;    while((c=getchar())<=32);    bool f=false;    if(c=='-')f=true,c=getchar();    for(x=0; c>32; c=getchar())x=x*10+c-'0';    if(f)x=-x;    return x;}template<class T>inline void write(T x){    if(x<0)putchar('-');    else if(x<10)putchar(x+'0');    else write(x/10),putchar(x%10+'0');}template<class T>inline void writeln(T x){    write(x);    puts("");}//--------IO template--------------------ll a[maxn];int n;int bin_ser(ll x)//二分查找{    int left=0,right=n-1;    while(left<=right)    {        int mid=(left+right)>>1;        if(a[mid]==x)return mid;        else if(x<a[mid])right=mid-1;        else left=mid+1;    }    return left-1;//没有的话，返回当前的位置}//不需要输入输出优化，只是自己这样写比较方便 int main(){    int m,i,j,k,t;    ll p;    read(n);read(p);    for(i=0;i<n;i++)read(a[i]);    if(p<0){write(0);return 0;}    sort(a,a+n);    i=0;    //while(a[i]*p<a[n-1])i++;write(n-i)///这样写只有一个case过不去，明显数据还是比较弱的    int ans=1;    for(i=0;i<n;i++)    {        ll x=p*a[i];        int index=bin_ser(x);//二分枚举        ans=max(index-i+1,ans);    }    writeln(ans);    return 0;}