【CodeForces-978D】解题报告（暴力，数学讨论）

原始题目

Almost Arithmetic Progression

  • time limit per test1 second
  • memory limit per test256 megabytes
  • inputstandard input
  • outputstandard output

Polycarp likes arithmetic progressions. A sequence [a1,a2,…,an] is called an arithmetic progression if for each i (1≤i<n) the value ai+1−ai is the same. For example, the sequences [42], [5,5,5], [2,11,20,29] and [3,2,1,0] are arithmetic progressions, but [1,0,1], [1,3,9] and [2,3,1] are not.

It follows from the definition that any sequence of length one or two is an arithmetic progression.

Polycarp found some sequence of positive integers [b1,b2,…,bn]. He agrees to change each element by at most one. In the other words, for each element there are exactly three options: an element can be decreased by 1, an element can be increased by 1, an element can be left unchanged.

Determine a minimum possible number of elements in b which can be changed (by exactly one), so that the sequence b becomes an arithmetic progression, or report that it is impossible.

It is possible that the resulting sequence contains element equals 0.

Input

The first line contains a single integer n (1≤n≤100000) — the number of elements in b.

The second line contains a sequence b1,b2,…,bn (1≤bi≤109).

Output

If it is impossible to make an arithmetic progression with described operations, print -1. In the other case, print non-negative integer — the minimum number of elements to change to make the given sequence becomes an arithmetic progression. The only allowed operation is to add/to subtract one from an element (can't use operation twice to the same position).

Examples

Input

4
24 21 14 10

Output

3

Input

2
500 500

Output

0

Input

3
14 5 1

Output

-1

Input

5
1 3 6 9 12

Output

1

Note

In the first example Polycarp should increase the first number on 1, decrease the second number on 1, increase the third number on 1, and the fourth number should left unchanged. So, after Polycarp changed three elements by one, his sequence became equals to [25,20,15,10], which is an arithmetic progression.

In the second example Polycarp should not change anything, because his sequence is an arithmetic progression.

In the third example it is impossible to make an arithmetic progression.

In the fourth example Polycarp should change only the first element, he should decrease it on one. After that his sequence will looks like [0,3,6,9,12], which is an arithmetic progression.

题目大意

给定一组数，对每一个数可以不操作，或者+1，或者-1（加减记录改变次数），问这一组数是否可以通过操作变为等差数列

  • 若可以，输出所需改变次数的最小值。
  • 若不可以，输出-1.

解题思路

对每种可能的gap遍历数据，更新需要更改的次数的最小值。 gap由前两项a[0],a[1]确定，共有9种可能。确定gap后开始从a[2]开始遍历数组，一旦出现不符合则跳出，全部符合则更新最小操作数。

解题代码

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#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <set>
#include <string>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
#define INF 0x3f3f3f3f
#define eps 1e-8
using namespace std;
const int maxn=1e5+5;
typedef long long ll;
ll a[maxn+10];
int ans; 
int n,m,t;
int main()
{
	while(~scanf("%d",&n))
	{
		ans=INF;
		for(int i=0;i<n;i++)
			scanf("%I64d",&a[i]);
		if(n==1)
		{
			printf("0\n");
			continue;
		}
		ll gap=a[1]-a[0];
		ll origin0=a[0],origin1=a[1],ogap=a[1]-a[0];
		
		//第一种情况
		for(int i=0;i<9;i++) 
		{
			int ccnt=0;
			if(i==0)
				a[0]=origin0-1,a[1]=origin1-1,ccnt=2;
			else if(i==1)
				a[0]=origin0,a[1]=origin1-1,ccnt=1;
			else if(i==2)
				a[0]=origin0+1,a[1]=origin1-1,ccnt=2;
			else if(i==3)
				a[0]=origin0-1,a[1]=origin1,ccnt=1;
			else if(i==4)
				a[0]=origin0,a[1]=origin1,ccnt=0;
			else if(i==5)
				a[0]=origin0+1,a[1]=origin1,ccnt=1;
			else if(i==6)
				a[0]=origin0-1,a[1]=origin1+1,ccnt=2;
			else if(i==7)
				a[0]=origin0,a[1]=origin1+1,ccnt=1;
			else if(i==8)
				a[0]=origin0+1,a[1]=origin1+1,ccnt=2;
			gap=a[1]-a[0];
			//开始遍历剩余
			int j=2;
			ll last=a[1];
			for(;j<n;j++)
			{
//				printf("before a[j]-last=%d\n",a[j]-last); 
				if(a[j]-last==gap)
				{
					last=a[j];
					continue;
				}
				else if(a[j]-last==gap+1)
				{
					last=a[j]-1;
					ccnt++;
					continue;
				}
				else if(a[j]-last==gap-1)
				{
					last=a[j]+1;
					ccnt++;
					continue;
				}
				else break;
			} 
			if(j==n) ans=min(ans,ccnt);
		}
		if(ans==INF) printf("-1\n");
		else printf("%d\n",ans);
	}
}

收获与反思

  • 想到可能的等差gap值。
  • 暴力遍历每种gap的可能。