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Description

一個長度為 $ n $ 的大數，用 $ S_1S_2S_3 \ldots S_n $表示，其中 $ S_i $ 表示數的第 $ i $ 位，$ S_1 $ 是數的最高位，告訴你一些限制條件，每個條件表示為四個數 $ (l_1, r_1, l_2, r_2) $，即兩個長度相同的區間，表示子串 $ S_{l_1}S_{l_1 + 1}S_{l_1 + 2} \ldots S_{r_1} $ 與 $ S_{l_2}S_{l_2 + 1}S_{l_2 + 2} \ldots S_{r_2} $ 完全相同。

比如 $ n = 6 $ 時，某限制條件 $ (l_1 = 1, r_1 = 3, l_2 = 4, r_2 = 6) $，那麼 $ 123123 \(、\)

\(n\le 10^5\)

Solution

可以想到，我們用 \(f_{i,j}\) 表示 \([i,i+2^j-1]\) 這一段區間，然後每次合併兩個區間就相當於合併 \(f_{l1,k},f_{l2,k}\) 以及 \(f_{r1-2^k+1,k},f_{r2-2^k+1,k}\) 。

然後我們還需要將兒子也下傳一下合併標記即可。

Code

#include <bits/stdc++.h>
using namespace std;

#define Int register int
#define mod 1000000007
#define MAXN 100005

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
template <typename T> inline void chkmax (T &a,T b){a = max (a,b);}
template <typename T> inline void chkmin (T &a,T b){a = min (a,b);}

int n,m,tot,lg[MAXN],f[MAXN][21],ls[MAXN * 21],rs[MAXN * 21],fa[MAXN * 21];

int findSet (int x){return fa[x] == x ? x : fa[x] = findSet (fa[x]);}
void unionSet (int x,int y){
	x = findSet (x),y = findSet (y);
	if (x == y) return ;
	fa[y] = x;
}

signed main(){
	read (n,m);
	for (Int i = 2;i <= n;++ i) lg[i] = lg[i >> 1] + 1;
	for (Int j = 0;(1 << j) <= n;++ j)
		for (Int i = 1;i + (1 << j) - 1 <= n;++ i)
			f[i][j] = ++ tot,fa[tot] = tot;
	for (Int j = 1;(1 << j) <= n;++ j)
		for (Int i = 1;i + (1 << j) - 1 <= n;++ i)
			ls[f[i][j]] = f[i][j - 1],rs[f[i][j]] = f[i + (1 << j - 1)][j - 1];
	for (Int i = 1;i <= m;++ i){
		int l1,r1,l2,r2;read (l1,r1,l2,r2);
		int k = lg[r1 - l1 + 1];
		unionSet (f[l1][k],f[l2][k]);
		unionSet (f[r1 - (1 << k) + 1][k],f[r2 - (1 << k) + 1][k]);
	}
	for (Int j = lg[n];j >= 1;-- j)
		for (Int i = 1;i + (1 << j) - 1 <= n;++ i){
			int t = findSet (f[i][j]);
			if (t == f[i][j]) continue;
			unionSet (ls[t],ls[f[i][j]]);
			unionSet (rs[t],rs[f[i][j]]);
		}
	int cnt = n,res = 9;
	for (Int i = 1;i <= n;++ i) if (findSet (f[i][0]) != f[i][0]) cnt --;
	for (Int i = 1;i < cnt;++ i) res = 1ll * res * 10 % mod;
	write (res),putchar ('\n');
	return 0;
}