loj#6074. 「2017 山东一轮集训 Day6」子序列(矩阵乘法 dp)

题意

题目链接

Sol

设\(f[i][j]\)表示前\(i\)个位置中，以\(j\)为结尾的方案数。

转移的时候判断一下\(j\)是否和当前位置相同

然后发现可以用矩阵优化，可以分别求出前缀积和逆矩阵的前缀积(这题的逆矩阵炒鸡好求)

这样就可以\(n*10^3\)

发现相邻两个矩阵只有一行不同，那么其他的可以直接copy。

就可以做到\(n*10^2\)了。

#include<bits/stdc++.h>

#define Pair pair<int, int>

#define MP(x, y) make_pair(x, y)

#define fi first

#define se second

#define LL long long

#define ull unsigned long long

#define Fin(x) {freopen(#x".in","r",stdin);}

#define Fout(x) {freopen(#x".out","w",stdout);}

using namespace std;

const int MAXN = 1e5 + 10, mod = 1e9 + 7, INF = 1e9 + 10;

const double eps = 1e-9;

template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}

template <typename A, typename B> inline int add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}

template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}

template <typename A, typename B> inline int mul(A x, B y) {return 1ll * x * y % mod;}

template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}

template <typename A> inline void debug(A a){cout << a << '\n';}

template <typename A> inline LL sqr(A x){return 1ll * x * x;}

template <typename A, typename B> inline LL fp(A a, B p, int md = mod) {int b = 1;while(p) {if(p & 1) b = mul(b, a);a = mul(a, a); p >>= 1;}return b;}

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

char s[MAXN];

int N, Q, a[MAXN];

struct Ma {

	int m[11][11];

	Ma() {

		memset(m, 0, sizeof(m));

	}

	void init() {

		for(int i = 0; i < 10; i++) m[i][i] = 1;

	}

	Ma operator * (const Ma &rhs) const {

		Ma ans;

		for(int i = 0; i < 10; i++)

			for(int j = 0; j < 10; j++) {

				__int128 tp = 0;

				for(int k = 0; k < 10; k++)

					tp += 1ll * m[i][k] * rhs.m[k][j];

				ans.m[i][j] = tp % mod;

			}

		return ans;

	}

}suf[MAXN], inv[MAXN];

int solve(int l, int r) {

	Ma tmp = suf[r] * inv[l - 1];

	int ans = 0;

	for(int i = 0; i < 10; i++) add2(ans, tmp.m[i][9]);

	return (ans - 1 + mod) % mod;

}

signed main() {

//	freopen("a.in", "r", stdin);

	scanf("%s", s + 1);

	N = strlen(s + 1);

	for(int i = 1; i <= N; i++) {

		a[i] = s[i] - 'a';

		suf[i].init(); inv[i].init();

		for(int j = 0; j < 10; j++) suf[i].m[a[i]][j] = 1;

		for(int j = 0; j < 10; j++) if(j != a[i]) inv[i].m[a[i]][j] = mod - 1;

	}

	suf[0].init(); inv[0].init();

	for(int i = 2; i <= N; i++) suf[i] = suf[i] * suf[i - 1];

	for(int i = 2; i <= N; i++) inv[i] = inv[i - 1] * inv[i];

	int Q = read();

	while(Q--) {

		int l = read(), r = read();

		cout << solve(l, r) << '\n';

	}

    return 0;

}

巴特西

loj#6074. 「2017 山东一轮集训 Day6」子序列(矩阵乘法 dp)

题意

Sol

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