HDU - 4990 Reading comprehension 【矩阵快速幂】

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=4990

题意
初始的ans = 0

给出 n, m

for i in 1 -> n

如果 i 为奇数
ans = (ans * 2 + 1) % m
反之
ans = ans * 2 % m

思路
如果我们只计算 偶数项 那么递推公式就是

ans[n] = 4 * ans[n - 2] + 2

如果 n 是偶数 那么刚好 就按这个公式推 第 n / 2 项
如果 n 是奇数 那么就是 第 【 n / 2 项 】 * 2 + 1

可以推知的矩阵为

然后矩阵快速幂

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1.0);
const double E = exp(1.0);
const double eps = 1e-8;

const int INF = 0x3f3f3f3f;
const int maxn = 1e2 + 5;
//const int MOD = 1e4;

int MOD;

struct Matrix
{
    ll a[2][2];
    Matrix () {}
    Matrix operator * (Matrix const &b)const
    {
        Matrix res;
        CLR(res.a, 0);
        for (int i = 0; i < 2; i++)
            for (int j = 0; j < 2; j++)
                for (int k = 0; k < 2; k++)
                    res.a[i][j] = (res.a[i][j] + this->a[i][k]  * b.a[k][j]) % MOD;
        return res;
    }
};

Matrix pow_mod(Matrix base, int n)
{
    Matrix ans;
    CLR(ans.a, 0);
    ans.a[0][1] = 1;
    while (n > 0)
    {
        if (n & 1)
            ans = ans * base;
        base = base * base;
        n >>= 1;
    }
    return ans;
}


int main()
{
    Matrix base;
    base.a[0][0] = 4;
    base.a[0][1] = 0;
    base.a[1][0] = 2;
    base.a[1][1] = 1;
    int n;
    while (~scanf("%d%d", &n, &MOD))
    {
        Matrix ans = pow_mod(base, n / 2);
        if (n & 1)
            printf("%lld\n", (ans.a[0][0] * 2 + 1) % MOD);
        else
            printf("%lld\n", ans.a[0][0] % MOD);
    }
}