1085. Perfect Sequence (25)

题目地址： http://www.patest.cn/contests/pat-a-practise/1085

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代码一：采用了二分搜索， 但是需要变化一下，很多细节需要注意

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <iterator>

using namespace std;

#define N 100001

int nn;
long int  p;
long int a
;

bool cmp(long int x, long int y)
{
return x < y;
}

int bsearch(int left, int right, long int n)
{
if (n >= a[right])//里面这里判断一下可以
return right;
while (left <= right)
{
int mid = (left + right) / 2;
if (a[mid] == n)
{
while (a[mid] == n)//这里相等 后不断的while == n
{
mid++;
if (mid > right)
return right;
}
return mid - 1;
}
else if (a[mid] < n)
{
if (a[mid + 1] > n) //这里是大于
{
return mid;
}
left = mid + 1;
}
else{
if (a[mid - 1] <= n) // 这里是小于等于
{
return mid - 1;
}
right = mid - 1;
}
}
}

int main()
{
//freopen("in", "r", stdin);
scanf("%ld%ld", &nn, &p);
int i, j;
for (i = 0; i < nn; i++)
{
scanf("%ld", &a[i]);
}
sort(a, a + nn, cmp);

int maxLen = 0;
for (i = 0; i < nn; i++)
{
long int m = a[i];
if (nn - i < maxLen) // 这里可以判断一下 nn - i
break;
int tmpL = 0;
long int M = m * p;
int pos = bsearch(i, nn - 1, M);
tmpL = pos - i + 1;
if (tmpL > maxLen)
{
maxLen = tmpL;
}
}

printf("%d\n", maxLen);
return 0;
}

ac代码二：需要加一个delta 优化

#include <stdio.h>
#include <iostream>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <iterator>

using namespace std;

#define N 100001

int nn;
long int  p;
long int a
;

bool cmp(long  int x, long int y)
{
return x < y;
}

int main()
{
// freopen("in", "r", stdin);
scanf("%ld%ld", &nn, &p);
int i, j;
for (i = 0; i < nn; i++)
{
scanf("%ld", &a[i]);
}
sort(a, a + nn, cmp);
int delta = 0;
int maxLen = 0;
for (i = 0; i < nn; i++)
{
if (nn - i + 1 < maxLen)
break;
int m = a[i];
int tmpL = 0;
// 这里是 i+delta - 1 不是 i+delta,自己做的时候搞错了，然后画了下，发现这个位置自己写错了
for (j = i+delta-1; j <= nn - 1; j++)
{
int M = a[j];
if (M > m * p)
{
tmpL = j - i;
break;
}
}
if (j == nn) // 这里到头了判断下
tmpL = nn - i;
if (tmpL > maxLen)
{
maxLen = tmpL;
delta = maxLen;
}
}

printf("%d\n", maxLen);
return 0;
}