time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Polycarp likes arithmetic progressions. A sequence?$[a1,a2,\dots ,an][a1,a2,\dots ,an]?is\; called\; an\; arithmetic\; progression\; if\; for\; each?ii?(1\le i<n1\le i<n)\; the\; value?ai+1?aiai+1?ai?is\; the\; same.\; For\; example,\; the\; sequences?[42][42],?[5,5,5][5,5,5],?[2,11,20,29][2,11,20,29]?and?[3,2,1,0][3,2,1,0]?are\; arithmetic\; progressions,\; but?[1,0,1][1,0,1],?[1,3,9][1,3,9]?and?[2,3,1][2,3,1]?are\; not.$

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

Polycarp found some sequence of positive integers?$[b1,b2,\dots ,bn][b1,b2,\dots ,bn].\; He\; agrees\; to\; change\; each\; element\; by\; at\; most\; one.\; In\; the\; other\; words,\; for\; each\; element\; there\; are\; exactly\; three\; options:\; an\; element\; can\; be\; decreased\; by?11,\; an\; element\; can\; be\; increased\; by?11,\; an\; element\; can\; be\; left\; unchanged.$

Determine a minimum possible number of elements in?$\mathrm{bb?which\; can\; be\; changed\; (by\; exactly\; one),\; so\; that\; the\; sequence?bb?becomes\; an\; arithmetic\; progression,\; or\; report\; that\; it\; is\; impossible.}$

It is possible that the resulting sequence contains element equals?$00.$

Input

The first line contains a single integer?$\mathrm{nn?(1\le n\le 100000)(1\le n\le 100000)?—\; the\; number\; of\; elements\; in?bb.}$

The second line contains a sequence?${\mathrm{b1,b2,\dots ,bnb1,b2,\dots ,bn?(1\le bi\le 109)(1\le bi\le 109).}}^{}$

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 Copy 4
24 21 14 10 output Copy 3 input Copy 2
500 500 output Copy 0 input Copy 3
14 5 1 output Copy -1 input Copy 5
1 3 6 9 12 output Copy 1 Note

In the first example Polycarp should increase the first number on?$\mathrm{11,\; decrease\; the\; second\; number\; on?11,\; increase\; the\; third\; number\; on?11,\; and\; the\; fourth\; number\; should\; left\; unchanged.\; So,\; after\; Polycarp\; changed\; three\; elements\; by\; one,\; his\; sequence\; became\; equals\; to?[25,20,15,10][25,20,15,10],\; which\; is\; an\; arithmetic\; progression.}$

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?$[0,3,6,9,12][0,3,6,9,12],\; which\; is\; an\; arithmetic\; progression.$

?

#include <iostream> #include <algorithm> #include <bits/stdc++.h> #define maxn 100005 using namespace std; typedef long long ll; int main() {int n,i,j,k;int a[maxn]={0};int b[maxn]={0};scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i]);}if(n==2||n==1){printf("0\n");return 0;}int minum=10000000;int flag=0;for(i=-1;i<=1;i++){for(j=-1;j<=1;j++){int cnt=abs(i)+abs(j);for(k=1;k<=n;k++){b[k]=a[k];}b[1]+=i;b[2]+=j;int x=b[1]-b[2];for(k=2;k<=n-1;k++){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){cnt++;b[k+1]-=1;}else{break;}if(k==n-1){minum=min(minum,cnt);flag=1;}}}}if(flag) printf("%d\n",minum);else printf("-1\n");return 0; }

　　

轉載于:https://www.cnblogs.com/zyf3855923/p/9039766.html

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