BZOJ2326 [HNOI2011]数学作业

最新推荐文章于 2018-12-18 20:15:00 发布

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本文介绍了一种使用矩阵快速幂解决特定数学问题的方法。通过构建一个3×3矩阵，并利用快速幂运算加速计算过程，该方法能够有效地求解10的幂次方问题。其时间复杂度为log²(n)。

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[Problem]

[Solution]

Just look at the range of N, and we know that it's about quick power.

Build a matrix of 3 * 3 like this:

10^x	1	1
0	1	1
0	0	1

and use it to multiply

Ans[i]

then, you can get the matrix

Ans[i + 1]

i + 1

That's the matrix construction of the problem.

You need to calculate each 10^x dividually. The time complex is log^2(n)

[Code]

#include <cstdio>
#include <algorithm>
#include <memory.h>

using namespace std;

typedef long long qw;

#ifdef WIN32
#define lld "%I64d"
#else
#define lld "%lld"
#endif

const int matsz = 3;

qw n, mod;

class matrix {
	public:
		int n;
		qw a[matsz][matsz];
		matrix() {
			memset(a, 0, sizeof(a));
			n = matsz;
		}
		void operator =(matrix x) {
			memcpy(a, x. a, sizeof(a));
		}
		matrix operator *(matrix x) {
			matrix s;
			for (int i = 0; i < n; i ++)
				for (int j = 0; j < n; j ++)
					for (int k = 0; k < n; k ++)
						(s. a[i][j] += a[i][k] * x. a[k][j]) %= mod;
			return s;
		}
		void print() {
			for (int i = 0; i < n; i ++, putchar(10))
				for (int j = 0; j < n; j ++)
					printf("%5d", (int)a[i][j]);
			putchar(10);
		}
};

matrix matPow(matrix a, qw n) {
	matrix s;
	bool f = 1;
	for (; n; n >>= 1, a = a * a)
		if (n & 1)
			(f) ? (s = a, f = 0) : (s = s * a, 0);
	return s;
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif

	matrix s;
	s. a[2][0] = 1;
	scanf(lld lld, &n, &mod);
	for (qw i = 1, m = 10; i <= n; i *= 10, (m *= 10) %= mod) {
		matrix a, r;
		a. a[0][0] = m;
		a. a[0][1] = 1;
		a. a[0][2] = 1;
		a. a[1][1] = 1;
		a. a[1][2] = 1;
		a. a[2][2] = 1;
		r = matPow(a, min(i * 10 - 1, n) - i + 1);
		s = r * s;
	}
	printf("%d\n", (int)s. a[0][0]);
}