poj 3233 矩阵快速幂-优快云博客

本文链接：https://blog.youkuaiyun.com/Fawkess/article/details/106999621

本文深入探讨了矩阵快速幂算法的实现细节，通过构造特定矩阵并利用快速幂运算，高效解决涉及多项式求值的问题。文章详细展示了算法的步骤，包括矩阵初始化、乘法运算以及快速幂过程，适用于处理大规模数据的计算。

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构造一个左上角是A+A²+A³ …的矩阵，然后去算就可以了

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace std;

typedef int Matrix[77][77];
Matrix tt;
int n, k, m;

void mul(const Matrix& a,const Matrix& b, Matrix &res) {
	for(int i = 0; i < 2 * n; i++)
		for (int j = 0; j < 2 * n; j++) {
			tt[i][j] = 0;
			for (int k = 0; k < 2 * n; k++)
				tt[i][j] += a[i][k] * b[k][j], tt[i][j] %= m;
		}
	for (int i = 0; i < 2 * n; i++)
		for (int j = 0; j < 2 * n; j++)
			res[i][j] = tt[i][j];
}

Matrix a,b;

int main() {
	scanf("%d%d%d", &n, &k, &m);
	k--;//这里也可以k不减，然后把a变成1，直接算就行了
	memset(a, 0, sizeof(a));
	memset(b, 0, sizeof(b));
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			scanf("%d", &a[i][j]), a[i][j] %= m, a[i][j + n] = a[i][j];
	for (int i = 0; i < n; i++)
		b[i][i] = 1;
	for (int i = 0; i < n; i++)
		for (int j = 0; j < n; j++)
			b[i + n][j] = b[i + n][j + n] = a[i][j];

	while (k) {
		if (k & 1)
			mul(a, b, a);
		mul(b, b, b);
		k >>= 1;
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++)
			printf("%d ", a[i][j]);
		printf("\n");
	}
		
	return 0;
}