B. Shift and Push

B. Shift and Push
time limit per test
1.0 s
memory limit per test
256 MB
input
standard input
output
standard output

Given two equally sized arrays A and B of size N. A is empty and B has some values.

You need to fill A with an element X such that X belongs to B.

The only operations allowed are:

1. Copy B[i] to A[i].

2. Cyclic shift B by 1 to the the right.

You need to minimise the number of operations.

Input

The first line contains a single positive integer N(1 ≤ N ≤ 106), denoting the size of the arrays.

Next line contains N space separated positive integers denoting the elements of the array B(1 ≤ B[i] ≤ 105).

Output

Output a single integer, denoting the minimum number of operations required.

Examples
Input
3
1 2 3
Output
5
Input
6
1 1 2 2 3 3
Output
10
Note

In the first test case:

We can have 5 steps as: fill first element, shift, fill second element, shift, fill third element.

Initially, A = [_, _, _], B = [1, 2, 3]

After step 1, A = [1, _, _], B = [1, 2, 3]

After step 2, A = [1, _, _], B = [3, 1, 2]

After step 3, A = [1, 1, _], B = [3, 1, 2]

After step 4, A = [1, 1, _], B = [2, 3, 1]

After step 5, A = [1, 1, 1], B = [2, 3, 1]

求每个数中相邻区间最大的最小的那个数。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <map>
 6 #include <set>
 7 #include <queue>
 8 #include <stack>
 9 #include <vector>
10 #define ll long long int
11 #define inf 0x3f3f3f3f
12 #define N 1000005
13 #define mem(a) memset(a,0,sizeof(a))
14 using namespace std;
15 int lowbit(int x){ return x&(-x);}
16 int a[N],b[N/10+5][5];
17 
18 int main(){
19   int n;
20   cin>>n;
21   for(int i = 1; i <= n; i ++){
22       scanf("%d", &a[i]);
23       if(!b[a[i]][0]) b[a[i]][1] = b[a[i]][2] = b[a[i]][3] = i;
24       else {
25           b[a[i]][2] = b[a[i]][3];
26           b[a[i]][3] = i;
27           b[a[i]][4] = max(b[a[i]][4], b[a[i]][3] - b[a[i]][2] - 1);
28       }
29       b[a[i]][0] ++;
30   }
31   int ans = inf;
32   for(int i = 1; i <= n; i ++) {
33       b[a[i]][4] = max(b[a[i]][4], n - b[a[i]][3] + b[a[i]][1] - 1);
34       ans = min(ans, b[a[i]][4] + n);
35   }
36   printf("%d\n",ans);
37   return 0;
38 }

 

posted @ 2017-10-01 23:48  #忘乎所以#  阅读(174)  评论(0)  编辑  收藏  举报