HDU5015 233 Matrix(矩阵高速幂)

HDU5015 233 Matrix(矩阵高速幂)

题目链接

题目大意：
给出n∗m矩阵，给出第一行a01， a02, a03 ...a0m (各自是233， 2333， 23333...), 再给定第一列a10， a10, a10, a10,...an0.矩阵中的每一个元素等于左边的加上上面的，求出anm.

解题思路：
先要依据矩阵元素的特征得出相乘的矩阵T, 然后就是求这个矩阵T的m次幂（这里就能够用矩阵高速幂），最后再和给定的第一列所形成的矩阵相乘，就能得到anm。
求矩阵T请參考

代码：

#include <cstdio>
#include <cstring>

typedef long long ll;

const int N = 15;
const ll MOD = 10000007;

ll A[N][N];
int B[N];
int n;
ll m;

struct Rec {

    ll v[N][N];

    Rec () { memset (v, 0, sizeof (v));}
    void init () {

        for (int i = 0; i < n + 2; i++)
            for (int j = 0; j < n + 2; j++)
                v[i][j] = A[i][j];
    }

    Rec operator * (const Rec &a) {

        Rec tmp;
        for (int i = 0; i < n + 2; i++)
            for (int j = 0; j < n + 2; j++) 
                for (int k = 0; k < n + 2; k++)
                    tmp.v[i][j] = (tmp.v[i][j] + (v[i][k] * a.v[k][j]) % MOD) % MOD;
        return tmp;
    }

    Rec operator *= (const Rec &a) {

        return *this = *this * a;
    }
}num;

void init () {

    memset (A, 0, sizeof (A));
    for (int i = 0; i < n + 1; i++) {
        A[i][0] = 10LL;
        A[i][n + 1] = 1LL;
    }

    A[n + 1][n + 1] = 1LL;
    for (int i = 1; i < n + 1; i++) 
        for (int j = 1; j <= i; j++) 
            A[i][j] = 1LL;
    B[0] = 23;
}

Rec f(ll m) {

    if (m == 1)
        return num;
    Rec tmp;
    tmp = f(m / 2);
    tmp *= tmp;
    if (m % 2)
        tmp *= num; 
    return tmp;
}

int main () {


    while (scanf ("%d%lld", &n, &m) != EOF) {

        for (int i = 1; i <= n; i++)
            scanf ("%d", &B[i]);

        init();
        B[n + 1] = 3;
        num.init ();

        num = f(m);

/*        for (int i = 0; i <= n + 1; i++) {
            for (int j = 0; j <= n + 1; j++)
                printf ("%lld ", num.v[i][j]);
            printf ("
");
        }*/

        ll ans = 0;
        for (int i = 0; i <= n + 1; i++) 
            ans = (ans + (num.v[n][i] * B[i]) % MOD) % MOD;
        printf ("%lld
", ans);
    }
    return 0;
}