HDU5015 233 Matrix(矩阵快速幂)

HDU5015 233 Matrix(矩阵快速幂)

题目链接

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

解题思路：
先要根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（这里就可以用矩阵快速幂），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请参考

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