AGC027 E ABBreviate

Solution

将’a’看成$1$，’b‘看成$2$。那么无论怎么变化，总和对于$3$的余数不变。
同时我们可以得到一个结论，对于一段$[l,r]$，如果最后这一段变成了某个字母，那么一定满足：

1. 总和膜$3$相等。
2. 至少有一对连续相等或者长度为一。

那么有了这个结论，考虑如何计算。对于一个$t$。首先其模$3$余数与$s$相同。我们每次一位一位匹配。每次贪心的向后找，直到能够凑成要凑的字母。这时候要注意，最后$s$可能还剩下一部分，由于要求同于，所以要保证剩下部分模$3$为$0$即可。这样可行的话，那么$t$就是可以变成的。
从而我们设$f[n]$表示这样贪心时$n$能够有多少种分法，每次贪心找一段即可。最后答案就是枚举剩下部分有多长。

code

如果任然不懂可以参见下面代码中的注释。

#include 
#define MOD 1000000007
#define INF 1061109567
#define bp __builtin_popcountll
#define pb push_back
#define in(s) freopen(s,"r",stdin);
#define out(s) freopen(s,"w",stdout);
#define fi first
#define se second
#define bw(i,r,l) for (int i=r-1;i>=l;i--)
#define fw(i,l,r) for (int i=l;i
#define fa(i,x) for (auto i:x)
using namespace std;
const int N = 1e5 + 5;
string s;
int n, dp[N], f[N][3], sum[N];
bool bad = 1;
void add(int &x, int y) {
	x += y;
	if (x >= MOD) x -= MOD;
}
signed main() {
	#ifdef BLU
	in("blu.inp");
	#endif
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
	cin >> s;
	int n = s.length();
	/*
	We can see greedily that always picking the earliest possible spots to create an a or b is good
	enough.
	dp[i]: Number of different substrings we can create from the first i characters of s
	Then pass onto 2 positions j1 and j2, so that i + 1 -> j1 can create 'a', and to j2 create 'b'.
	For a, psum[j1] - psum[i] must mod 3 remainder 1. So we can know the remainder psum[j1] should
	have, and this can be easily found by a supporting array.
	*/
	s = " " + s;
	memset(sum, 0, sizeof sum);
	fw (i, 1, n + 1) sum[i] = sum[i - 1] + (s[i] == 'a' ? 1 : 2);
	memset(f, -1, sizeof f);
	/*
	f[i][j]: The first index k that considering the segment from i + 1 -> k, its sum MOD 3 is equal to
	j. Also, there are at least 2 adjancent characters that are equal (or the length is 1)
	*/
	fw (i, 1, n) if (s[i] == s[i + 1]) bad = 0;
	if (bad) {
		cout << "1";
		return 0;
	}
	bw (i, n + 1, 0) {
		fw (j, 0, 3) f[i][j] = f[i + 1][j];
		if (i) f[i][sum[i] % 3] = i;
	}
	memset(dp, 0, sizeof dp);
	dp[0] = 1;
	fw (i, 0, n + 1) {
		fw (j, 1, 3) {
			int k = (j + sum[i]) % 3;
			k = f[i][k];
			if (k != -1) {
				add(dp[k], dp[i]);
				
				bool flag = k == i + 1;
				for (int d = i + 2;d <= k;d++){
					if (s[d] == s[d-1]) flag |= 1;
				}
				assert(flag);
			}
		}
	}
	int ans = 0;
	fw (i, 1, n + 1) {
		if ( (sum[n] - sum[i]) % 3 == 0) add(ans , dp[i]);
		printf("%d\n",dp[i]);
	}
	cout << ans;
	return 0;
}