Hdu5015 233 matrix (matrix fast power)

Hdu5015 233 matrix (matrix fast power)

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Question:
GivenN?MMatrix, indicating the first line A01, A02, A03... a0m (233,233 3, 23333 ...), specify the first column A10, A10, A10, A10 ,... in the an0.0. each element in the matrix is equal to the above on the left side, and ANM is obtained.

Solution:
First, the multiplication Matrix T should be obtained based on the features of the matrix element, and then the M power of the matrix T is obtained (here we can use the matrix Rapid power ), then multiply the matrix formed by the given first column to obtain the ANM.
For Matrix T, see

Code:

#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;}

Hdu5015 233 matrix (matrix fast power)