1085. Perfect Sequence (25) -- 二分法

题目地址

https://www.patest.cn/contests/pat-a-practise/1085

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 82 3 20 4 5 1 6 7 8 9

Sample Output:

8

ac代码(二分法思路)

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <iostream>#include <string>#include <vector>#include <queue>#include <algorithm>#include <sstream>#include <list>#include <stack> #include <map> #include <set> #include <iterator> using namespace std;// freopen("in.txt", "r", stdin);#define INF 0x7ffffffftypedef long long int LL;LL n, p;vector<LL> v;// 找到第一个大于val的数 返回下标int bsearch(int left, int right ,LL val){    if(left > right)        return -1;    if(val > v[right])        return right + 1;    if(val < v[left])        return -1;    int tmpL = left;    int tmpR = right;    while(left < right) // left == right 将退出    {        int mid = (right - left) / 2 + left;        if(val == v[mid])        {            int j = mid;            while(v[j] <= val)                j ++;            return j; // 第一个大于 val的数的下标        }else if(val > v[mid])        {            left = mid + 1;        }else{            right = mid - 1;        }    }    int j = min(left, right);    int k;    for(k = j ; j < tmpR;j ++)    {        if(v[j] > val)        {            break;        }    }    return k;}int main( ){    //freopen("in.txt","r",stdin);    while(scanf("%lld%lld", &n, &p) != EOF)    {        v.resize(n);        for(int i=0;i<n;i++)        {            scanf("%lld", &v[i]);        }        sort(v.begin(), v.end());        int l = 0;        int r = n - 1;        if(v[r] <= v[l] * p)        {            printf("%d\n", n);        }else{            int maxLen = 1;            for(int i=0;i<n-1;i++)            {                LL val = v[i] * p; //                 int t = bsearch(i + 1, n-1 , val);                if(t != -1)                {                    if(t - i > maxLen)                    {                        maxLen = t - i;                    }                }            }            printf("%d\n", maxLen);        }    }    return 0;}

超时代码

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <iostream>#include <string>#include <vector>#include <queue>#include <algorithm>#include <sstream>#include <list>#include <stack> #include <map> #include <set> #include <iterator> using namespace std;// freopen("in.txt", "r", stdin);#define INF 0x7ffffffftypedef long long int LL;LL n, p;vector<LL> v;int main( ){    //freopen("in.txt","r",stdin);  while(scanf("%lld%lld", &n, &p) != EOF)  {    v.resize(n);    for(int i=0;i<n;i++)    {      scanf("%lld", &v[i]);    }    sort(v.begin(), v.end());    int l = 0;    int r = n - 1;    if(v[r] <= v[l] * p)    {      printf("%lld\n", n);    }else{      int maxLen = -1;      for(int i=0;i<n-1;i++)      {        if(n - 1 - i + 1 < maxLen)          break;        int cnt = 1;        for(int j = i+1;j < n;j++)        {          if(v[j] <= v[i] * p)          {            cnt ++;          }        }        if(cnt > maxLen)        {          maxLen = cnt;        }      }      printf("%d\n", maxLen);    }  }  return 0;}