hihoCoder 1133-二分·二分查找之k小数 #1133 : 二分·二分查找之k小数 时间限制:10000ms 单点时限:1000ms 内存限制:256MB 描述 在上一回里我们知道Nettle在玩《艦これ》，Nettle的镇守府有很多船位，但船位再多也是有限的。Nettle通过捞船又出了一艘稀有的船，但是已有的N(1≤N≤1,000,000)个船位都已经有船了。所以Nettle不得不把其中一艘船拆掉来让位给新的船。Nettle思考了很久，决定随机选择一个k，然后拆掉稀有度第k小的船。 已知每一艘船都有自己的稀有度，Nettle现在把所有船的稀有度值告诉你，希望你能帮他找出目标船。 提示：非有序数组的二分查找之二 输入 第1行：2个整数N,k。N表示数组长度， 第2行：N个整数，表示a[1..N]，保证不会出现重复的数，1≤a[i]≤2,000,000,000。 输出 第1行：一个整数t，表示t在数组中是第k小的数，若K不在数组中，输出-1。 样例输入 10 4 1732 4176 2602 6176 1303 6207 3125 1 1011 6600 样例输出 1732
给定一个无序序列，问序列中第k小的数是哪个。简单粗暴的方法是排序直接取a[k]，复杂度为O(nlgn)；还有一种O(lgn)的方法，采用了和hihoCoder 1128-二分·二分查找类似的方法，将序列不断二分成小于某个数和大于某个数的两部分，直到某一部分长度为k。 完整代码如下： [cpp] #include<iostream> #include<cstdio> #include<vector> using namespace std; int n, k; vector<int> a; int GetNumberK(int p, int q) { int m = a[p], i = p, j = q; while (i < j) { while (i < j&&a[j] > m) j–; a[i] = a[j]; while (i < j&&a[i] < m) i++; a[j] = a[i]; } if (k == i + 1)//(1) return m; else if (k < i + 1) return GetNumberK(p, i – 1); else { //k = k – (i – p + 1);//(2)，因为(1)处判等的时候i取p，但分割之后p并没有从0开始，所以k还是k return GetNumberK(i + 1, q); } } int main() { //freopen("input.txt", "r", stdin); scanf("%d %d", &n, &k); a.resize(n); for (int i = 0; i < n; i++) scanf("%d", &a[i]); if (k > n||k < 1) printf("-1\n"); else printf("%d\n", GetNumberK(0, n – 1)); return 0; } [/cpp] 本代码提交AC，用时69MS，内存3MB。 需要注意一点，代码中(2)不能出现，虽然递归在[i+1,q]内部查找第k – (i – p + 1)小的数，但是(1)处判等的时候，使用的i（即p）是相对[0,n-1]长度的，所以这里的k不需要剪短]]>
所以结果为$$-1modn=n-1$$。
当n=2和4时，需要特殊处理。完整代码如下：
[cpp]
#include<iostream>
#include<cstdio>
using namespace std;
bool IsPrime(int x)
{
for (int i = 2; i*i <= x; i++)
if (x%i == 0)
return false;
return true;
}
int main()
{
int t, n;
scanf("%d", &t);
while (t–)
{
scanf("%d", &n);
if (n == 2)
printf("1n");
else if (n == 4)
printf("2n");
else if (IsPrime(n))
printf("%dn", n – 1);
else
printf("0n");
}
return 0;
}
[/cpp]
本代码提交AC，用时967MS，内存1576K。]]>
This record defines a permutation P as follows: P(1) = 4, P(2) = 1, P(3) = 5, etc.
What is the value of the expression P(P(1))? It’s clear, that P(P(1)) = P(4) = 2. And P(P(3)) = P(5) = 3. One can easily see that if P(n) is a permutation then P(P(n)) is a permutation as well. In our example (believe us)
It is natural to denote this permutation by P2(n) = P(P(n)). In a general form the defenition is as follows: P(n) = P1(n), Pk(n) = P(Pk-1(n)). Among the permutations there is a very important one — that moves nothing:
It is clear that for every k the following relation is satisfied: (EN)k = EN. The following less trivial statement is correct (we won’t prove it here, you may prove it yourself incidentally): Let P(n) be some permutation of an N elements set. Then there exists a natural number k, that Pk = EN. The least natural k such that Pk = EN is called an order of the permutation P.
The problem that your program should solve is formulated now in a very simple manner: “Given a permutation find its order.”
Input
In the first line of the standard input an only natural number N (1 <= N <= 1000) is contained, that is a number of elements in the set that is rearranged by this permutation. In the second line there are N natural numbers of the range from 1 up to N, separated by a space, that define a permutation — the numbers P(1), P(2),…, P(N).
Output
You should write an only natural number to the standard output, that is an order of the permutation. You may consider that an answer shouldn’t exceed 109.
Sample Input
5
4 1 5 2 3
Sample Output
6
Source
Ural State University Internal Contest October’2000 Junior Session
Another sample test case.