ural 1252. Sorting the Tombstones

1252. Sorting the Tombstones

Time limit: 1.0 second
Memory limit: 64 MB
There is time to throw stones and there is time to sort stones…
An old desolate cemetery is a long dismal row of nameless tombstones There are N tombstones of various shapes. The weights of all the stones are different. People have decided to make the cemetery look more presentable, sorting the tombstone according to their weight. The local custom allows to transpose stones if there are exactly K other stones between them.

Input

The first input line contains an integer N (1 ≤ N ≤ 130000). Each of the next N lines contains an integer X, the weight of a stone in grams (1 ≤ X ≤ 130000).

Output

The output should contain the single integer — the maximal value of K (0 ≤ K < N), that makes possible the sorting of the stones according to their weights.

Sample

inputoutput
5
30
21
56
40
17

1
Problem Author: Alexey Lakhtin
Problem Source: Open collegiate programming contest for student teams, Ural State University, March 15, 2003
Difficulty: 417
 
题意：给一个序列n个数，如果只能交换相距为k的两个数，然后能够通过这种交换使原序列有序，那么这个k是符合性质的，求最大的k，输出k-1。
分析：如果一开始就是有序的，答案就是n-1。
否则，考虑一个合法的k，
再考虑一个数到应该去的位置的距离x，
显然有k整除于x，即   k|x
那么这个k就是每个数到应该去的位置的距离的gcd。
注意有序既可以是从小到大，又可以是从大到小。
 
  1 /**
  2 Create By yzx - stupidboy
  3 */
  4 #include <cstdio>
  5 #include <cstring>
  6 #include <cstdlib>
  7 #include <cmath>
  8 #include <deque>
  9 #include <vector>
 10 #include <queue>
 11 #include <iostream>
 12 #include <algorithm>
 13 #include <map>
 14 #include <set>
 15 #include <ctime>
 16 #include <iomanip>
 17 using namespace std;
 18 typedef long long LL;
 19 typedef double DB;
 20 #define MIT (2147483647)
 21 #define INF (1000000001)
 22 #define MLL (1000000000000000001LL)
 23 #define sz(x) ((int) (x).size())
 24 #define clr(x, y) memset(x, y, sizeof(x))
 25 #define puf push_front
 26 #define pub push_back
 27 #define pof pop_front
 28 #define pob pop_back
 29 #define ft first
 30 #define sd second
 31 #define mk make_pair
 32 
 33 inline int Getint()
 34 {
 35     int Ret = 0;
 36     char Ch = ' ';
 37     bool Flag = 0;
 38     while(!(Ch >= '0' && Ch <= '9'))
 39     {
 40         if(Ch == '-') Flag ^= 1;
 41         Ch = getchar();
 42     }
 43     while(Ch >= '0' && Ch <= '9')
 44     {
 45         Ret = Ret * 10 + Ch - '0';
 46         Ch = getchar();
 47     }
 48     return Flag ? -Ret : Ret;
 49 }
 50 
 51 const int N = 130010;
 52 int n, arr[N], order[N];
 53 int ans;
 54 
 55 inline void Input()
 56 {
 57     scanf("%d", &n);
 58     for(int i = 0; i < n; i++) scanf("%d", &arr[i]);
 59 }
 60 
 61 inline int Gcd(int a, int b)
 62 {
 63     if(b) return Gcd(b, a % b);
 64     else return a;
 65 }
 66 
 67 inline int Work(int *arr, int *order, int n)
 68 {
 69     int ret = 0;
 70     for(int i = 0; i < n; i++)
 71     {
 72         int idx = lower_bound(order, order + n, arr[i]) - order;
 73         int delta = abs(i - idx);
 74         ret = Gcd(ret, delta);
 75     }
 76     return ret;
 77 }
 78 
 79 inline void Solve()
 80 {
 81     ans = 0;
 82     for(int i = 0; i < n; i++) order[i] = arr[i];
 83     sort(order, order + n);
 84     int t1 = Work(arr, order, n);
 85     for(int i = 0; i < n; i++) arr[i] = order[i] = - arr[i];
 86     sort(order, order + n);
 87     int t2 = Work(arr, order, n);
 88     if(!t1 || !t2) ans = n;
 89     else ans = max(t1, t2);
 90     
 91     printf("%d\n", ans - 1);
 92 }
 93 
 94 int main()
 95 {
 96     freopen("a.in", "r", stdin);
 97     Input();
 98     Solve();
 99     return 0;
100 }
View Code

 

posted @ 2015-12-26 10:48  yanzx6  阅读(164)  评论(0)  编辑  收藏  举报