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 }
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