codeforces - 978D【思维】
1 second
256 megabytes
standard input
standard output
Polycarp likes arithmetic progressions. A sequence [a1,a2,…,an][a1,a2,…,an] is called an arithmetic progression if for each ii (1≤i<n1≤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,…,bn][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 11 , an element can be increased by 11 , an element can be left unchanged.
Determine a minimum possible number of elements in 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 .
The first line contains a single integer nn(1≤n≤100000)(1≤n≤100000) — the number of elements in bb .
The second line contains a sequence b1,b2,…,bnb1,b2,…,bn(1≤bi≤109)(1≤bi≤109) .
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).
4
24 21 14 10
3
2
500 500
0
3
14 5 1
-1
5
1 3 6 9 12
1
In the first example Polycarp should increase the first number on 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.
解题思路:等差数列的公差相等, 所以前两个数就可以确定公差,记录后面的数满足公差需要多少步即可,O(6 * n)的复杂度。
附ac代码:
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <string>
5 #include <cmath>
6 #include <string>
7 #include <iostream>
8 #include <map>
9 #include <queue>
10 #include <stack>
11 #include <cstdlib>
12 const int maxn = 3 * 1e5 + 10;
13 const int inf = 0x3f3f3f3f;
14
15 using namespace std;
16 typedef long long ll;
17 const ll mod = 1e9 + 7;
18 int nu[maxn];
19 int tem[maxn];
20 queue<int>q;
21
22 int main(int argc, const char * argv[]) {
23 int n;
24 scanf("%d", &n);
25 for(int i = 0; i < n; ++i)
26 {
27 scanf("%d", &nu[i]);
28 }
29 if(n <= 2)
30 {
31 puts("0");
32 return 0;
33 }
34 int i, j, k;
35 int ans = inf;
36 for(i = -1; i <= 1; ++i)
37 {
38 for(j = -1; j <= 1; ++j)
39 {
40 tem[0] = nu[0] + i;
41 tem[1] = nu[1] + j;
42 int d = tem[1] - tem[0];
43 int cnt = abs(i) + abs(j);
44 for(k = 2; k < n; ++k)
45 {
46 int u = nu[k] - tem[k - 1];
47 tem[k] = d + tem[k - 1];
48 if(u == d) continue;
49 if(abs(u - d) <= 1)
50 {
51 cnt++;
52 // printf("%d ", tem[k]);
53 }
54 else break;
55 }
56 if(k == n)
57 {
58 ans = min(cnt, ans);
59 }
60 }
61 }
62 if(ans == inf) puts("-1");
63 else
64 printf("%d\n", ans);
65 return 0;
66 }
codeforces - 978D【思维】的更多相关文章
- Codeforces 424A (思维题)
Squats Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Statu ...
- Codeforces 1060E(思维+贡献法)
https://codeforces.com/contest/1060/problem/E 题意 给一颗树,在原始的图中假如两个点连向同一个点,这两个点之间就可以连一条边,定义两点之间的长度为两点之间 ...
- Queue CodeForces - 353D (思维dp)
https://codeforces.com/problemset/problem/353/D 大意:给定字符串, 每一秒, 若F在M的右侧, 则交换M与F, 求多少秒后F全在M左侧 $dp[i]$为 ...
- codeforces 1244C (思维 or 扩展欧几里得)
(点击此处查看原题) 题意分析 已知 n , p , w, d ,求x , y, z的值 ,他们的关系为: x + y + z = n x * w + y * d = p 思维法 当 y < w ...
- CodeForces - 417B (思维题)
Crash Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit Status ...
- CodeForces - 417A(思维题)
Elimination Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Submit ...
- CodeForces 625A 思维
题意是说一个人喝酒 有两种办法 买塑料瓶的 a块钱 喝了就没了 或者是买玻璃瓶的b块钱 喝完还能卖了瓶子c块钱 求最多能喝多少瓶 在开始判断一次 a与b-c的关系 即两种方式喝酒的成本 如果a< ...
- Vladik and Complicated Book CodeForces - 811B (思维实现)
Vladik had started reading a complicated book about algorithms containing n pages. To improve unders ...
- The Contest CodeForces - 813A (思维)
Pasha is participating in a contest on one well-known website. This time he wants to win the contest ...
随机推荐
- 前端开发好帮手,eslint配置全知道
eslint让人又爱又恨,原因在于它的默认配置非常严格,动则一个小提示就直接报错不给运行.而在开发调试的过程中,我们想时时得到运行效果,它的严格又很烦. 在安装eslint后,我们可以在package ...
- SparkStreaming和Kafka基于Direct Approach如何管理offset实现exactly once
在之前的文章<解析SparkStreaming和Kafka集成的两种方式>中已详细介绍SparkStreaming和Kafka集成主要有Receiver based Approach和Di ...
- AWS IoT Greengrass是什么?V1和V2版本及其差异
AWS IoT Greengrass Greengrass主要是用于边缘计算或者机器学习有关,对于详细了解请阅读结尾处的官方文档,文档内容也较为丰富. 目录 AWS IoT Greengrass ...
- Tensorflow-基础使用
Tensorflow基本概念 使用图(graphs)来表示计算任务 在被称之为会话(Session)的上下文(context)中执行图 使用tensor表示数据 通过变量(Variable)维护状态 ...
- selenium八大元素定位方法
1.ID定位 可以根据元素的id来定位属性,id是当前整个HTML页面中唯一的,所以可以通过id属性来唯一定位一个元素,是首选的元素定位方式.(动态ID不做考虑) # 导入webdriver和By f ...
- 有状态(Stateful)应用的容器化 - 云+社区 - 腾讯云 https://cloud.tencent.com/developer/article/1020178
有状态(Stateful)应用的容器化 - 云+社区 - 腾讯云 https://cloud.tencent.com/developer/article/1020178
- win api 窗口操作-窗口置顶与寻找与激活
https://docs.microsoft.com/en-us/windows/win32/api/winuser/nf-winuser-setwindowpos https://docs.micr ...
- 你的隐私数据真的安全吗之memset()使用分析
我们在实际编程中,需要保存许多私有数据,例如:密码.密钥等等.所以,我们需要经常在使用完这些私有数据后,清除内存使用踪迹,以防止被潜在的入侵者获得这些数据.这篇文章中,我们讨论使用memset()函数 ...
- ShowDoc,APIDoc,可道云API,语雀-适合IT企业的文档工具
ShowDoc,APIDoc,可道云API,语雀-适合IT企业的文档工具 一.ShowDoc官方文档及说明 1.1 它可以用来做什么 1.2 它都有些什么功能 1.3 使用在线的ShowDoc 1.4 ...
- c++复习笔记(4)
这一篇是另一篇各种琐碎东西的笔记. 类型转换可以通过类型转换函数,或者构造函数来实现.但是一般来说类型转换指的是类型转换函数. 类型转换函数不需要声明输出类型(因为输出类型是固定的),也没有参数,同时 ...