HDU-4565 So Easy! 公式化简

题意：非常简单，就是求一个表达式的最后结果。

解法：http://blog.csdn.net/magic____/article/details/9021169

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long LL;

LL a, b, n, m;

struct Matrix {
    int r, c;
    int a[2][2];
    void unit() {
        memset(a, 0, sizeof (a));
        for (int i = 0; i < r; ++i) {
            a[i][i] = 1;
        }
    }
    void init() {
        memset(a, 0, sizeof (a));
    }
    Matrix operator * (const Matrix & ot) const {
        Matrix ret;
        ret.r = r, ret.c = ot.c;
        ret.init();
        for (int i = 0; i < ret.r; ++i) {
            for (int j = 0; j < ret.c; ++j) {
                for (int k = 0; k < c; ++k) {
                    ret.a[i][j] += a[i][k] * ot.a[k][j];
                    ret.a[i][j] %= m;
                }
            }
        }
        return ret;
    }
    void show() {
        puts("");
        for (int i = 0; i < r; ++i) {
            for (int j = 0; j < c; ++j) {
                printf("%I64d ", a[i][j]);
            }
            puts("");
        }
        puts("");
    }
};

Matrix _pow(Matrix a, LL b) {
    Matrix ret;
    ret.r = ret.c = 2;
    ret.unit();
    while (b) {
        if (b & 1) {
            ret = ret * a;
        }
        b >>= 1;
        a = a * a;
    }
    return ret;
}

int main() {
    while (scanf("%I64d %I64d %I64d %I64d", &a, &b, &n, &m) != EOF) {
        Matrix A, B;
        A.init();
        A.r = 2, A.c = 1;
        A.a[0][0] = (2*a*a+2*b) % m, A.a[1][0] = (2*a) % m;
        if (n == 1) {
            printf("%I64d\n", A.a[1][0]);
            continue;
        } else if (n == 2) {
            printf("%I64d\n", A.a[0][0]);
            continue;
        }
        B.r = B.c = 2;
        B.a[0][0] = (2*a)%m, B.a[0][1] = (((b-a*a)%m)+m)%m;
        B.a[1][0] = 1, B.a[1][1] = 0;
        A = _pow(B, n-2) * A;
        printf("%I64d\n", A.a[0][0]);
    }
    return 0;
}