52 lines
1.4 KiB
C++
52 lines
1.4 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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//矩阵快速幂
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typedef long long LL;
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const int MOD = 1e9 + 7;
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const int N = 110;
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LL n, k; //本题数据范围很大，用int直接wa哭了
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LL C[N][N], A[N][N], B[N][N];
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//矩阵乘法
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void mul(LL C[][N], LL A[][N], LL B[][N]) {
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//为什么非得要开一个临时的tmp来存结果呢，直接用C确实不行，原理不清楚
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static LL tmp[N][N];
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memset(tmp, 0, sizeof tmp);
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for (LL i = 0; i < n; i++)
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for (LL j = 0; j < n; j++)
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for (LL k = 0; k < n; k++) {
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tmp[i][j] += A[i][k] * B[k][j] % MOD; //矩阵乘法
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tmp[i][j] %= MOD;
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}
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memcpy(C, tmp, sizeof tmp);
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}
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void qmi() {
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//将结果矩阵初始化为单位矩阵
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memset(B, 0, sizeof B);
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for (int i = 0; i < n; i++) B[i][i] = 1;
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//其实就是把整数快速幂修改为矩阵快速幂
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while (k) { //二进制快速幂
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if (k & 1) mul(B, B, A); // 联想一下整数快速幂
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mul(A, A, A); // base 翻倍
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k >>= 1;
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}
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}
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int main() {
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cin >> n >> k;
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//输入原始矩阵
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for (int i = 0; i < n; i++)
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for (int j = 0; j < n; j++)
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cin >> A[i][j];
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//计算矩阵快速幂
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qmi();
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//输出
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++)
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cout << B[i][j] << " ";
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cout << endl;
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}
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return 0;
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} |