题目描述
https://www.luogu.com.cn/problem/P1939
Solution
我们来考虑如何构造矩阵
好了，考虑完了
时间复杂度 $O(9T\log n)$
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| #include <iostream>
#include <cstring>
#include <cstdio>
#define cM const Matrix
using namespace std;
const int p = 1000000007;
inline int add(int x, int y) { return x + y >= p ? x + y - p : x + y; }
int n;
struct Matrix {
int a[4][4], n;
Matrix() { memset(a, 0, sizeof a); n = 3; }
inline void setone() { for (int i = 1; i <= n; ++i) a[i][i] = 1; }
friend Matrix operator * (cM &u, cM &v) {
Matrix w; int n = w.n;
for (int k = 1; k <= n; ++k)
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= n; ++j)
w.a[i][j] = add(w.a[i][j], 1ll * u.a[i][k] * v.a[k][j] % p);
return w;
}
} ;
Matrix pow(Matrix x, int n) {
Matrix s; s.setone();
for (; n; n >>= 1) {
if (n & 1) s = s * x;
x = x * x;
}
return s;
}
void work() {
cin >> n; Matrix a, b; a.a[1][1] = a.a[2][1] = a.a[3][1] = 1;
b.a[1][1] = b.a[1][3] = b.a[2][1] = b.a[3][2] = 1;
if (n <= 3) return (void) puts("1");
printf("%d\n", (pow(b, n - 3) * a).a[1][1]);
}
int main() {
int T; cin >> T;
while (T--) work();
return 0;
}
|