AGC027 E - ABBreviate

---

[TOC]

#题目链接
AGC027 E - ABBreviate
#题解
神仙啊
建议查看https://img.atcoder.jp/agc027/editorial.pdf
定义a = 1,b = 1发现在%3的情况下所有变换的相等的
性质:一个字符串，能变成字符c的条件是val[a] == val[c]并且a中有一个可以变换的位置
只需要判断这种变换不经过ababab这种就好了
那我们要求得就把字符串划分为k段，第i段的val值和第i字符的val值相等
f_i表示前i个看做一段的方案数
#代码

#include<cstdio> 
#include<cctype> 
#include<cstring> 
#include<algorithm> 
#define gc getchar() 
#define pc putchar
#define LL long long
inline int read() { 
	int x = 0,f = 1; 
	char c = getchar(); 
	while(c < '0' || c > '9') c = gc; 
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar(); 
	return x * f; 
}
void print(LL x) { 
	if(x < 0) { 
		pc('-'); 
		x = -x; 
	} 
	if(x >= 10) print(x / 10); 
	pc(x % 10 + '0'); 
} 
const int maxn = 100007; 
char t[maxn]; 
int n,s[maxn]; 
int f[maxn]; 
const int mod = 1e9 + 7; 
int main() { 
	scanf("%s",t + 1); 
	n = strlen(t + 1); 
	bool flag = false ; 
	for(int i = 2;i <= n;++ i) if(t[i] == t[i - 1])flag = true; 
	if(!flag) { 
		puts("1");  
		return 0; 
	} 
	int now = 0,pre[3] = {0,-1,-1}; 
	f[0] = 1; 
	for(int i = 1;i <= n;++ i) { 
		now = (now + 2 - (t[i] == 'a')) % 3; 
		if(!now && i < n) f[i] = 1; 
		++ now,now %= 3; 
		if(pre[now] != - 1) (f[i] += f[pre[now]]) %= mod;  
		++ now,now %= 3; 
		if(pre[now] != -1) (f[i] += f[pre[now]]) %= mod;
		++ now,now %= 3; 
		pre[now] = i; 
	} 
	print(f[n]); 
}