CodeForces 1238D AB-string

cf题面

## 解题思路

比赛过程中好像完全没有想法来着，打完比赛，群里分享的做法也并没有看懂……一个月后的今天终于靠官方题解把题补了。

可以用间接法，总的子串数量减去不好的子串数量就是好的子串数量。而不好的子串就4种——

• ABB...BB
• BAA...AA
• AA...AAB
• BB...BBA

其他情况都是好的子串。证明的话分类讨论一下就好

• 不好的字母在子串中间
  • 该字母左右的字母是否相等
• 不好的字母在子串边上
  • 子串中是否存在另一个和该字母相同的字母

官方题解这么说的——

Let's call a character (t_i) in string (t_1t_2…t_k) is bad if there is no such palindrome (t_lt_{l+1}…t_r) that (lleqslant ileqslant r). Any character in substring (t_2t_3…t_{k−1}) is good. It can be proven as follows. If (t_i=t_i+1) or (t_i=t_i−1) then (t_i) belong to a palindrome of length 2. If (t_i ot = t_{i+1}) and (t_i≠t_{i−1}) then (t_i) belong to a palindrome (t_{i−1}…t_{i+1}).

So only characters (t1) and (tk) can be bad. But at the same time character (t1) is bad if there is no character (ti) such that (i>1) and (t_i=t_1). It is true because substring (t_1t_2…t_i) is palindrome (index (i) is minimum index such that (t_i=t_1)).

So, there are only 4 patterns of bad strings:

1. ABB…BBABB…BB;
2. BAA…AABAA…AA;
3. AA…AABAA…AAB;
4. BB…BBABB…BBA;

All that remains is to count the number of substrings of this kind.

源代码

写的有点丑，本来可以一个循环完事的。

#include<cstdio>

long long n;
char s[300005];
int cnt[300005]={1},seg;
long long ans;
int main()
{
	scanf("%lld",&n);
	scanf("%s",s);
	int pos=0;
	for(int i=1;i<n;i++)
	{
		if(s[i]==s[pos]) cnt[seg]++;
		else
		{
			seg++;
			cnt[seg]=1;
			pos=i;
		}
	}
	//for(int i=0;i<=seg;i++) printf("%d ",cnt[i]);
	ans=n*(n-1)>>1;
	for(int i=1;i<=seg;i++)// cnt[i] and cnt[i-1]
	{
		ans-=cnt[i-1]+cnt[i]-1;
	}
	printf("
%lld
",ans);
	return 0;
}