[AHOI2004]数字迷阵

瞪眼观察题。发现行列都跟斐波那契数列有关。

如果能求出第一列就好了。

发现第一列的数都是由若干个$2$和若干$3$的和加$1$。

发现第$fib[i]$行有$fib[i-2]$个3和$fib[i-3]$个2。

其他行呢？

将行号拆分成若干项斐波那契数列的和，发现对应的$3$、$2$的数目也是对应的和。

然后万事大吉了。

距离$AC$还只需要会矩阵快速幂。

感性理解代码。

#include <bits/stdc++.h>

using namespace std;

#define re register
#define rep(i, a, b) for (re int i = a; i <= b; ++i)
#define repd(i, a, b) for (re int i = a; i >= b; --i)
#define For(i, a, b, s) for (re int i = a; i <= b; s)
#define maxx(a, b) a = max(a, b)
#define minn(a, b) a = min(a, b)
#define LL long long
#define INF (1 << 30)

inline int read() {
    int w = 0, f = 1; char c = getchar();
    while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();
    while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();
    return w * f;
}

const int MAX = 1000000000;

struct Matrix {
    int a, b, c, d;
    // | a b |
    // | c d |
} f, g;

int fib[100], x, y, P;
int a, b;

Matrix operator * (Matrix a, Matrix b) {
    Matrix res;
    res.a = (a.a * b.a + a.b * b.c) % P;
    res.b = (a.a * b.b + a.b * b.d) % P;
    res.c = (a.c * b.a + a.d * b.c) % P;
    res.d = (a.c * b.b + a.d * b.d) % P;
    return res;
}

int main() {
    fib[1] = fib[2] = 1;
    int X = x = read()-1; y = read(), P = read();
    int i;
    for (i = 3; fib[i-1] <= x; i++) fib[i] = fib[i-1] + fib[i-2];
    while (i--)
        if (fib[i] <= x) x -= fib[i], a = (a + 3 * fib[i-1]) % P, b = (b + 2 * fib[i-2]) % P;
    a = (a+b+1) % P, b = (a*2ll+(LL)X*P-X) % P;
    if (y < 3) printf("%d", y == 1 ? a : b);
    else {
        g.a = g.b = g.c = 1; g.d = 0;
        f.a = f.d = 1; f.b = f.c = 0;
        y -= 2;
        while (y) {
            if (y & 1) f = f * g;
            g = g * g;
            y >>= 1;
        }
        printf("%d", (f.b * a + f.a * b) % P);
    }
    return 0;
}