Recursive sequence(矩阵快速幂)

这就是矩阵的魅力吗，爱了爱了

题

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<math.h>
#include<set>
#define N 100
#define LL long long
using namespace std;
const LL mod = 2147493647;
struct Mat
{LL m[10][10];
};
Mat a, e;
Mat Mul(Mat x, Mat y)
{Mat c;for (int i = 1; i <= 7; i++) {for (int j = 1; j <= 7; j++) {c.m[i][j] = 0;}}for (int i = 1; i <= 7; i++) {for (int j = 1; j <= 7; j++) {for (int k = 1; k <= 7; k++) {c.m[i][j] = c.m[i][j] % mod + x.m[i][k] * y.m[k][j] % mod;}}}return c;
}
Mat pow(Mat x, LL y)
{Mat ans;for (int i = 1; i <= 7; i++){for (int j = 1; j <= 7; j++){if (i == j)ans.m[i][i] = 1;else ans.m[i][j] = 0;}}while (y) {if (y & 1) ans = Mul(ans, x);x = Mul(x, x);y >>= 1;}return ans;
}
int main()
{int t;scanf("%d", &t);LL n, f1, f2;Mat f;for (int i = 1; i <= 7; i++){for (int j = 1; j <= 7; j++)f.m[i][j] = 0;}f.m[1][1] = 1, f.m[1][2] = 1;f.m[2][1] = 2;f.m[3][1] = 1, f.m[3][3] = 1;f.m[4][3] = 4, f.m[4][4] = 1;f.m[5][3] = 6, f.m[5][4] = 3, f.m[5][5] = 1;f.m[6][3] = 4, f.m[6][4] = 3, f.m[6][5] = 2, f.m[6][6] = 1;f.m[7][3] = 1, f.m[7][4] = 1, f.m[7][5] = 1, f.m[7][6] = 1, f.m[7][7] = 1;while (t--){scanf("%lld%lld%lld", &n, &f1, &f2);e.m[1][1] = f2, e.m[1][2] = f1, e.m[1][3] = 81, e.m[1][4] = 27, e.m[1][5] = 9, e.m[1][6] = 3, e.m[1][7] = 1;Mat ans = pow(f, n - 2);LL sum = 0;for (int i = 1; i <= 7; i++){sum += e.m[1][i] * ans.m[i][1] % mod;sum %= mod;}printf("%lld\n", sum);} return 0;
}

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