codeforc 603-A. Alternative Thinking(规律)
2 seconds
256 megabytes
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO) in the form of a binary string of length n. Each character of Kevin's string represents Kevin's score on one of the n questions of the olympiad—'1' for a correctly identified cow and '0' otherwise.
However, all is not lost. Kevin is a big proponent of alternative thinking and believes that his score, instead of being the sum of his points, should be the length of the longest alternating subsequence of his string. Here, we define an alternating subsequence of a string as a not-necessarily contiguous subsequence where no two consecutive elements are equal. For example, {0, 1, 0, 1}, {1, 0, 1}, and {1, 0, 1, 0} are alternating sequences, while {1, 0, 0} and {0, 1, 0, 1, 1} are not.
Kevin, being the sneaky little puffball that he is, is willing to hack into the USAICO databases to improve his score. In order to be subtle, he decides that he will flip exactly one substring—that is, take a contiguous non-empty substring of his score and change all '0's in that substring to '1's and vice versa. After such an operation, Kevin wants to know the length of the longest possible alternating subsequence that his string could have.
The first line contains the number of questions on the olympiad n (1 ≤ n ≤ 100 000).
The following line contains a binary string of length n representing Kevin's results on the USAICO.
Output a single integer, the length of the longest possible alternating subsequence that Kevin can create in his string after flipping a single substring.
8
10000011
5
2
01
2
In the first sample, Kevin can flip the bolded substring '10000011' and turn his string into '10011011', which has an alternating subsequence of length 5: '10011011'.
In the second sample, Kevin can flip the entire string and still have the same score.
http://codeforces.com/problemset/problem/603/A;
题意:给你一串01串。
然后你可以选其中的一个连续子串,然后反转一下,就是0变1,1变0,然后求01相互间个的子序列最长是多少。
如果你选一串子串,反转,改变的只是子串两端的关系,子串内部的结构不会改变,比如1001001,变为0110110,那么内部的01还是原来的个数,
那么好了,改变前后01最多增加2个,可以假设原来已经成01或10 的不变(贪心原则)那么只要把那些11,和00,的改变,
计算前后两个相等的个数,然后和2比较取小的就行,再加上原来的01个数,就行了.
1 #include<stdio.h>
2 #include<algorithm>
3 #include<stdlib.h>
4 #include<iostream>
5 using namespace std;
6 char a[100005];
7 int main(void)
8 {
9 int i,j,k,p,q;
10 while(scanf("%d",&k)!=EOF)
11 {
12 scanf("%s",a);
13 int sum=0;
14 int cnt=0;
15 for(i=1; a[i]!='\0'; i++)
16 {
17 if(a[i]==a[i-1])
18 {
19 sum++;
20 }
21 else cnt++;
22
23 }
24 int r=min(2,sum);
25 printf("%d\n",cnt+r+1);
26
27 }
28 return 0;
29
30 }
codeforc 603-A. Alternative Thinking(规律)的更多相关文章
- Codeforces 603A - Alternative Thinking - [字符串找规律]
题目链接:http://codeforces.com/problemset/problem/603/A 题意: 给定一个 $01$ 串,我们“交替子序列”为这个串的一个不连续子序列,它满足任意的两个相 ...
- Alternative Thinking 找规律
Kevin has just recevied his disappointing results on the USA Identification of Cows Olympiad (USAICO ...
- cf Round 603
A.Alternative Thinking(思维) 给出一个01串,你可以取反其中一个连续子串,问取反后的01子串的最长非连续010101串的长度是多少. 我们随便翻一个连续子串,显然翻完之后,对于 ...
- 代码的坏味道(9)——异曲同工的类(Alternative Classes with Different Interfaces)
坏味道--异曲同工的类(Alternative Classes with Different Interfaces) 特征 两个类中有着不同的函数,却在做着同一件事. 问题原因 这种情况往往是因为:创 ...
- hdu1452 Happy 2004(规律+因子和+积性函数)
Happy 2004 题意:s为2004^x的因子和,求s%29. (题于文末) 知识点: 素因子分解:n = p1 ^ e1 * p2 ^ e2 *..........*pn ^ en 因子 ...
- Codeforces Round #384 (Div. 2) B. Chloe and the sequence(规律题)
传送门 Description Chloe, the same as Vladik, is a competitive programmer. She didn't have any problems ...
- ACM/ICPC 之 DP解有规律的最短路问题(POJ3377)
//POJ3377 //DP解法-解有规律的最短路问题 //Time:1157Ms Memory:12440K #include<iostream> #include<cstring ...
- HDU 5795 A Simple Nim 打表求SG函数的规律
A Simple Nim Problem Description Two players take turns picking candies from n heaps,the player wh ...
- 在sqlserver中做fibonacci(斐波那契)规律运算
--利用sqlserver来运算斐波那契规律 --利用事物与存储过程 declare @number intdeclare @A intdeclare @B intdeclare @C int set ...
随机推荐
- 单片机ISP、IAP和ICP几种烧录方式的区别
单片机ISP.IAP和ICP几种烧录方式的区别 玩单片机的都应该听说过这几个词.一直搞不太清楚他们之间的区别.今天查了资料后总结整理如下. ISP:In System Programing,在系统编程 ...
- CentOS6忘记root密码如何重置
CentOS6忘记root密码,如何重置密码 ① 重启服务器,按"e"键进入修改系统开机项界面 ② 选择内核项,按"e"进入其中进行配置 在文件内 ...
- URLDNS分析
学习了很久的Java基础,也看了很多的Java反序列化分析,现在也来分析学习哈最基础的URLDNS反序列化吧. Java反序列化基础 为了方便数据的存储,于是乎有了现在的Java序列化于反序列化.序列 ...
- Spark(二十一)【SparkSQL读取Kudu,写入Kafka】
目录 SparkSQL读取Kudu,写出到Kafka 1. pom.xml 依赖 2.将KafkaProducer利用lazy val的方式进行包装, 创建KafkaSink 3.利用广播变量,将Ka ...
- 容器之分类与各种测试(三)——queue
queue是单端队列,但是在其实现上是使用的双端队列,所以在queue的实现上多用的是deque的方法.(只要用双端队列的一端只出数据,另一端只进数据即可从功能上实现单端队列)如下图 例程 #incl ...
- ping (网络诊断工具)
Ping是Windows.Unix和Lnix系统下的一个命令,ping也属于一个通信协议,是TCP/IP协议的一部分,利用Ping命令可以检查网络是否连通,可以很好地帮助我们分析和判定网络故障.应用格 ...
- android TabLayout设置选项卡之间的距离无效已解决
根据下面的链接设置完距离后无法生效 https://www.jb51.net/article/131304.htm layout <com.google.android.material.tab ...
- C++ 类型转换(C风格的强制转换):
转https://www.cnblogs.com/Allen-rg/p/6999360.html C++ 类型转换(C风格的强制转换): 在C++基本的数据类型中,可以分为四类:整型,浮点型,字符型, ...
- maven项目install时忽略执行test
1.在项目所在文件夹根目录使用maven命令打包时: <!-- 不执行单元测试,也不编译测试类 --> mvn install -Dmaven.test.skip=true 或 <! ...
- leetcode,两个排序数组的中位数
先上题目描述: 给定两个大小为 m 和 n 的有序数组 nums1 和 nums2 . 请找出这两个有序数组的中位数.要求算法的时间复杂度为 O(log (m+n)) . 你可以假设 nums1 和 ...