[洛谷P4838]P哥破解密码

最新推荐文章于 2025-08-09 16:37:48 发布

转载最新推荐文章于 2025-08-09 16:37:48 发布 · 158 阅读

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原文链接：http://www.cnblogs.com/Memory-of-winter/p/9881997.html

文章标签：

#c/c++

本文介绍了一种求解特定条件下01串数量的高效算法。通过定义结尾为1、连续两个1及结尾为0的串个数，利用矩阵快速幂优化递推过程，解决了求长度为n的01串中不包含连续三个1的问题。

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题目大意：求长度为$n$的$01$串中，没有连续至少$3$个$1$的串的个数

题解：令$a_1$为结尾一个$1$的串个数，$a_2$为结尾两个$1$的串的个数，$b$为结尾是$0$的串的个数。$a_1=b,a_2=a_1,b=a_1+a_2+b$。

卡点：无

C++ Code：

#include <cstdio>
const int mod = 19260817;
int Tim, n;
inline void up(int &a, int b) {if ((a += b) >= mod) a -= mod;}
struct matrix {
	#define M 3
	int s[M][M];
	inline matrix() {
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) s[i][j] = 0;
		}
	}
	inline void init() {
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) s[i][j] = 0;
		}
		s[0][1] = s[0][2] = 1;
		s[1][2] = 1;
		s[2][0] = s[2][2] = 1;
	}
	inline void init(int a) {
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) s[i][j] = 0;
		}
		s[0][2] = 1;
	}
	inline int getans() {
		int res = 0;
		for (int i = 0; i < M; i++) up(res, s[0][i]);
		return res;
	}
	inline friend matrix operator * (const matrix &lhs, const matrix &rhs) {
		matrix res;
		for (int i = 0; i < M; i++) {
			for (int j = 0; j < M; j++) {
				for (int k = 0; k < M; k++) {
					up(res.s[i][j], static_cast<long long>(lhs.s[i][k]) * rhs.s[k][j] % mod);
				}
			}
		}
		return res;
	}
} base, ans;

int solve(int n) {
	base.init();
	ans.init(1);
	for (; n; n >>= 1, base = base * base) if (n & 1) ans = ans * base;
	return ans.getans();
}
int main() {
	scanf("%d", &Tim);
	while (Tim --> 0) {
		scanf("%d", &n);
		printf("%d\n", solve(n));
	}
	return 0; 
}

转载于:https://www.cnblogs.com/Memory-of-winter/p/9881997.html