AGC027 E - ABBreviate

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目录

• [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]);
}