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 }
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 }
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 }