UVA - 10870 Recurrences 【矩阵快速幂】

题目链接

https://odzkskevi.qnssl.com/d474b5dd1cebae1d617e6c48f5aca598?v=1524578553

题意

给出一个表达式 算法 f(n)

思路

n 很大 自然想到是 矩阵快速幂

那么问题就是 怎么构造矩阵

我们想到的一种构造方法是

n = 2 时

n = 3 时

然后大概就能够发现规律了吧 。。

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 <list>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a, b) memset(a, (b), sizeof(a))
#define pb push_back
#define bug puts("***bug***");
#define fi first
#define se second
#define stack_expand #pragma comment(linker, "/STACK:102400000,102400000")
#define syn_close   ios::sync_with_stdio(false);cin.tie(0);
#define sp system("pause");
//#define bug 
//#define gets gets_s

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;
typedef pair <double, int> pdi;

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 + 10;
const int MOD = 142857;

int d, n, m;

ll a[20], b[20];

struct Matrix
{
    ll a[20][20];
    Matrix() {}
    Matrix operator * (Matrix const &b)const
    {
        Matrix res;
        CLR(res.a, 0);
        for (int i = 0; i < d; i++)
            for (int j = 0; j < d; j++)
                for (int k = 0; k < d; k++)
                    res.a[i][j] = (res.a[i][j] + this->a[i][k] * b.a[k][j]) % m;
        return res;
    }
};

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

int main()
{
    while (scanf("%d %d %d", &d, &n, &m) && (d || n || m))
    {
        for (int i = 0; i < d; i++)
            scanf("%lld", &a[i]);
        for (int i = 0; i < d; i++)
            scanf("%lld", &b[i]);
        if (n <= d)
        {
            printf("%lld\n", b[n - 1] % m);
            continue;
        }
        Matrix ans;
        for (int i = 0; i < d; i++)
            for (int j = 0; j < d; j++)
                ans.a[i][j] = b[d - j - 1];
        ans = pow_mod(ans, n - d);
        printf("%lld\n", ans.a[0][0]);
    }
    return 0;
}