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HDU 5015 233 Matrix

2016-03-05 15:04 447 查看
矩阵快速幂。

首先得到这个式子:



依据这个,就可以构造矩阵。



#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

long long const MOD = 10000007;
int n, m;
long long a[20];

struct Matrix
{
long long A[20][20];
int R, C;
Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
Matrix c;
memset(c.A, 0, sizeof(c.A));
int i, j, k;
for (i = 1; i <= R; i++)
for (j = 1; j <= b.C; j++)
for (k = 1; k <= C; k++)
c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j]) % MOD) % MOD;
c.R = R; c.C = b.C;
return c;
}

void init()
{
memset(X.A, 0, sizeof X.A);
memset(Y.A, 0, sizeof Y.A);
memset(Z.A, 0, sizeof Z.A);

Z.R = n + 2; Z.C = 1;
for (int i = 1; i <= n; i++) Z.A[i][1] = a[i]; Z.A[n + 1][1] = 23; Z.A[n + 2][1] = 3;

X.R = n + 2; X.C = n + 2;
for (int i = 1; i <= n; i++) for (int j = 1; j <= i; j++) X.A[i][j] = 1;
for (int i = 1; i <= n + 1; i++) X.A[i][n + 1] = 10;
for (int i = 1; i <= n + 2; i++) X.A[i][n + 2] = 1;

Y.R = n + 2; Y.C = n + 2;
for (int i = 1; i <= n + 2; i++) Y.A[i][i] = 1;
}

void read()
{
for (int i = 1; i <= n; i++)
{
scanf("%lld", &a[i]);
a[i] = a[i] % MOD;
}
}

void work()
{
while (m)
{
if (m % 2 == 1) Y = Y*X;
m = m >> 1;
X = X*X;
}
Z = Y*Z;

printf("%lld\n", Z.A
[1]);
}

int main()
{
while (~scanf("%d%d", &n, &m))
{
read();
init();
work();
}
return 0;
}
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