Almost Arithmetic Progression

A. 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
inputCopy
4
24 21 14 10
outputCopy
3
inputCopy
2
500 500
outputCopy
0
inputCopy
3
14 5 1
outputCopy
-1
inputCopy
5
1 3 6 9 12
outputCopy
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.

思路：
通过调整前两个数，求得公差，依次对后面数操作，满足公差要求，若能一直操作结束，则满足题意，注意，需要在9次调整中选最小的，9次调整，可以通过循环得到。

package 一月十三;

import java.util.Scanner;

public class almostArithmeticprogressoon {

	private static final int  maxn=100005;

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int n,i,j,k;
		int[] a=new int[maxn];
		int[] b=new int[maxn];
		Scanner sc=new Scanner(System.in);
		n=sc.nextInt();
		for ( i = 1; i <=n; i++) {
			a[i]=sc.nextInt();
		}
		sc.close();
		//一个或两个数直接返回
		if(n==2||n==1) {
			System.out.println(0);
			return;
		}
		//最小值不会大于n
		int minum=n;
		//如果n==0，返回-1；否则返回minum;
		int flag=0;
		//暴力求解，外层循环i,代表对第一个元素的操作；
		for(i=-1;i<=1;i++) {
			//代表对第二元素操作，外循环三次，内循环三次，总共3*3次；
			for(j=-1;j<=1;j++) {
				//cnt代表对前两个元素的操作数量，i，j==1或-1；均进行1次，0进行次；
				int cnt=Math.abs(i)+Math.abs(j);
				
				//复制数组位b，对操作，因为要对a 操作9次，必须保证a不被修改
				for(k=0;k<=n;k++) {
					b[k]=a[k];
				}
				//对b 进行操作，在i，j取不同值
				b[1]+=i;
				b[2]+=j;
				//x代表差值
				int x=b[1]-b[2];
				//从第三项与第二项差开始，一直到最后两项的差值；
				for(k=2;k<n;k++) {
					//如果前一项减去后一项加1后等于x;对x进行加1操作
					if(b[k]-(b[k+1]+1)==x) {
						b[k+1]+=1;
						cnt++;
					}
					else if(b[k]-b[k+1]==x) {
						
					}
					else if(b[k]-(b[k+1]-1)==x) {
						b[k+1]-=1;
						cnt++;
					}
					//+，-不变都不满足，进行结束for循环
					else {
						break;
					}
					
				}
				//循环结束后，判断，如果k==n,则未执行break;说明满足题意### 注意，是n 不是n-1; 因为最后一次会加1；
				//若能达到n ，flag置为1；
				if(k==n) {
					minum=Math.min(minum,cnt);flag=1;
				}

			
			
				
			}
		}
		//若为1，则存在，输出，否则为0；
		//当n ==1||n==2时，已经在前面结束程序
		if(flag==1) {
			System.out.println(minum);
		}
		else System.out.println(-1);
	}

}