AtCoder Beginner Contest 200 F Minflip Summation

洛谷传送门

AtCoder 传送门

显然的策略：选择全部 \(0\) 段变成 \(1\)，或选择全部 \(1\) 段变成 \(0\)。

归纳可得一般性的结论：设字符串中 \(s_i \ne s_{i+1}\) 的位置数为 \(k\)，答案为 \(\left\lceil\frac{k}{2}\right\rceil\)。

因为在模意义下不能上取整，考虑记 \(k\) 的奇偶性（这样统计答案的时候就知道答案是 \(\frac{k}{2}\) 还是 \(\frac{k+1}{2}\)），然后 dp。设 \(f_{i,0/1,0/1}\) 为前 \(i\) 个字符中，\(k\) 的奇偶性，\(s_i\) 填的数，\(k\) 的总和。类似地设 \(g_{i,0/1,0/1}\) 表示方案数。转移是 trivial 的。

因为 \(K \le 10^9\)，考虑把转移写成 \(8 \times 8\) 的矩阵乘法形式，预处理第一个 copy 的 dp 值，做一遍矩阵快速幂就行。

时间复杂度 \(O(8^3 (n + \log K))\)。

code

// Problem: F - Minflip Summation
// Contest: AtCoder - KYOCERA Programming Contest 2021（AtCoder Beginner Contest 200）
// URL: https://atcoder.jp/contests/abc200/tasks/abc200_f
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 100100;
const ll mod = 1000000007;
const ll inv2 = (mod + 1) / 2;

struct matrix {
	ll a[8][8];
	matrix() {
		mems(a, 0);
	}
} ma[maxn], I;

inline matrix operator * (const matrix &a, const matrix &b) {
	matrix res;
	for (int i = 0; i < 8; ++i) {
		for (int j = 0; j < 8; ++j) {
			for (int k = 0; k < 8; ++k) {
				res.a[i][j] = (res.a[i][j] + a.a[i][k] * b.a[k][j] % mod) % mod;
			}
		}
	}
	return res;
}

inline matrix qpow(matrix a, ll p) {
	matrix res = I;
	while (p) {
		if (p & 1) {
			res = res * a;
		}
		a = a * a;
		p >>= 1;
	}
	return res;
}

int n, m, id[2][2][2];
ll f[maxn][2][2], g[maxn][2][2];
char s[maxn];

void solve() {
	for (int i = 0; i < 8; ++i) {
		I.a[i][i] = 1;
	}
	scanf("%s%d", s + 1, &m);
	n = strlen(s + 1);
	for (int i = 0; i < 2; ++i) {
		for (int j = 0; j < 2; ++j) {
			for (int k = 0; k < 2; ++k) {
				id[i][j][k] = i * 4 + j * 2 + k;
			}
		}
	}
	if (s[1] == '0' || s[1] == '?') {
		g[1][0][0] = 1;
	}
	if (s[1] == '1' || s[1] == '?') {
		g[1][0][1] = 1;
	}
	for (int i = 2; i <= n; ++i) {
		for (int j = 0; j < 2; ++j) {
			for (int x = 0; x < 2; ++x) {
				for (int y = 0; y < 2; ++y) {
					if (s[i] != '?' && s[i] != '0' + y) {
						continue;
					}
					int nj = j ^ (x ^ y);
					f[i][nj][y] = (f[i][nj][y] + f[i - 1][j][x]) % mod;
					if (x ^ y) {
						f[i][nj][y] = (f[i][nj][y] + g[i - 1][j][x]) % mod;
					}
					g[i][nj][y] = (g[i][nj][y] + g[i - 1][j][x]) % mod;
				}
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j < 2; ++j) {
			for (int x = 0; x < 2; ++x) {
				for (int y = 0; y < 2; ++y) {
					if (s[i] != '?' && s[i] != '0' + y) {
						continue;
					}
					int nj = j ^ (x ^ y);
					++ma[i].a[id[0][j][x]][id[0][nj][y]];
					if (x ^ y) {
						++ma[i].a[id[1][j][x]][id[0][nj][y]];
					}
					++ma[i].a[id[1][j][x]][id[1][nj][y]];
				}
			}
		}
	}
	matrix a = I, b;
	for (int i = 1; i <= n; ++i) {
		a = a * ma[i];
	}
	for (int i = 0; i < 2; ++i) {
		for (int j = 0; j < 2; ++j) {
			b.a[0][id[0][i][j]] = f[n][i][j];
			b.a[0][id[1][i][j]] = g[n][i][j];
		}
	}
	matrix res = b * qpow(a, m - 1);
	ll ans = 0;
	for (int i = 0; i < 2; ++i) {
		ans = (ans + (res.a[0][id[0][1][i]] + res.a[0][id[1][1][i]]) % mod * inv2 % mod) % mod;
		ans = (ans + res.a[0][id[0][0][i]] * inv2 % mod) % mod;
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}