一本通1642：【例 2】Fibonacci 第 n 项

#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
typedef signed long long ll;
struct node {
    ll p[3][3];
    node() {
        memset(p, 0, sizeof(p));
    }
};
int n;
ll M;
node A;
node plus(node a, node b) {
    node c;

    for (int i = 1; i <= 2; i++)
        for (int j = 1; j <= 2; j++)
            for (int k = 1; k <= 2; k++)
                c.p[i][j] += a.p[i][k] * b.p[k][j], c.p[i][j] %= M;

    return c;
}
node plus1(node a, node b, int n, int m, int p) {
    node c;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= p; j++)
            for (int k = 1; k <= m; k++)
                c.p[i][j] += a.p[i][k] * b.p[k][j], c.p[i][j] %= M;

    return c;
}
node mi(int n) {
    if (n == 1)
        return A;

    node t = mi(n / 2);

    if (n % 2 == 1)
        return plus(plus(t, t), A);
    else
        return plus(t, t);
}
int main() {
    A.p[1][1] = 1;
    A.p[1][2] = 1;
    A.p[2][1] = 1;
    A.p[2][2] = 0;
    scanf("%d %llu", &n, &M);
    node c = mi(n - 2), ha;
    ha.p[1][1] = 1;
    ha.p[2][1] = 1;
    c = plus1(c, ha, 2, 2, 1);
    printf("%llu\n", c.p[1][1] % M);
    return 0;
}