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 \(l\leqslant i\leqslant r\). 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 \(t_i\not = t_{i+1}\) 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:
- ABB…BBABB…BB;
- BAA…AABAA…AA;
- AA…AABAA…AAB;
- 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("\n%lld\n",ans);
return 0;
}