A1085. Perfect Sequence

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

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 long long num[100001];
 6 bool cmp(long long a, long long b){
 7     return  a < b;
 8 }
 9 int binSearch(long long num[], int low, int high, long long m, long long p){
10     long long ans = p * m;
11     int mid = high;
12     while(low < high){
13         mid = (low + high) / 2;
14         if(num[mid] > ans){
15             high = mid;
16         }else low = mid + 1;
17     }
18     return low;
19 }
20 int main(){
21     int N, p, index, ans = -1;
22     scanf("%d%d", &N, &p);
23     for(int i = 0; i < N; i++)
24         scanf("%lld", &num[i]); 
25     sort(num, num + N, cmp); 
26     for(int i = 0; i < N; i++){
27         index = binSearch(num, i + 1, N, num[i], p);
28         index--;
29         if(index - i + 1 > ans){
30             ans = index - i + 1;
31         }
32     }
33     printf("%d", ans); 
34     cin >> N;
35     return 0;
36 }
View Code

总结:

1、题意:给出一个数列和一个数字p,求满足M <= mp 的最长子数列, M与m分别为最大最小值。先对原数列排序一下,对比条件,显然M越大、m越小可以得到的子数列长度越长。因此可以从排好序的数列的两端往中间搜索答案,但是如果暴力二重循环会出现超时。因此,只能从左端递增遍历m, 而右端的M改用二分搜索,将复杂度变为 n*logn。

2、二分搜索M,目标是找到序列中尽可能靠右的且满足 M <= mp 的M, 即最后一个满足所给条件的M。这个可以转化为搜索第一个满足 M > mp 的num[i]。 则num[i - 1]即为最后一个满足M <= mp的M。

3、本题还可以使用两点法做,即使用两个指针,一遍扫描。如果a[j] <= a[i]* p成立,那么在[i, j]之内的任意k,都有a[k] <= a[i] *p成立,基于这一点设置i、j两个指针,j先递增,发现不等式不成立之后,i再向前走一步,如此反复。双指针法不仅可以一头一尾,也可以在同方向使用。代码如下:

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 using namespace std;
 5 long long num[100001];
 6 bool cmp(long long a, long long b){
 7     return  a < b;
 8 }
 9 
10 int main(){
11     int N, p, index, ans = -1;
12     scanf("%d%d", &N, &p);
13     for(int i = 0; i < N; i++)
14         scanf("%lld", &num[i]); 
15     sort(num, num + N, cmp); 
16     int i = 0, j = 0;
17     while(i < N && j < N){
18         long long temp = num[j] - num[i] * p;
19         if(temp <= 0){
20             ans = max(ans, j - i);
21             j++;
22         }else{
23             i++;
24         }
25     }
26     printf("%d", ans + 1); 
27     cin >> N;
28     return 0;
29 }
View Code

 

4、二分搜索:

1)搜索满足某条件的元素。  

int binSearch(int num[], int low, int high, int x){
    int mid;
    while(low <= high){ //当搜索区间为空时结束
        mid = low + (high - low) / 2;
        if(num[mid] == x)
            return mid;
        else if(num[mid] > x)
            high = mid - 1;
        else low = mid + 1;
    }    
    return -1;
}

2)搜索第一个满足某条件的元素,在这个情况下要注意,被排好序的序列一定是从左到右先不满足条件,然后满足;另外,传入的high = N时,在搜索不到满足条件的数时,会返回N(假想N位置有数字); 返回的是low而不是mid。

/*返回第一个大于等于x的元素。num[N]数组应从小到大排序。
如果传入的high = N,则当序列中所有元素都小于x时,会返回N */
int binSearch(int num[], int low, int high, int x){
    int mid;
    while(low < high){  //当low == high时结束搜索
        mid = low + (high - low) / 2;
        if(num[mid] >= x)
            high = mid;  //mid有可能就是结果,所以区间上限不能漏掉mid
        else low = mid + 1;
    }    
    return low;  //返回low而不是mid
}

 

posted @ 2018-02-03 09:20  ZHUQW  阅读(134)  评论(0编辑  收藏  举报