HDU 5015 2014 ACM/ICPC Asia Regional Xi'an Online 233 Matrix

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当天比赛没想到是矩阵快速幂， 弱渣一个===

题意不用说， 构造矩阵， A(i, j) = A(i-1, j) + A(i, j-1) = ............递推一下就知道了， 所以我们可以按照下面的构造

代码

#include <stdio.h>
#include <cmath>
#include <cstring>
using namespace std;
#define N 15
#define MOD 10000007
long long n, m, a[15];

void matric_mul(long long a[][N], long long b[][N]); //矩阵相乘
void makemat(long long mat[][N]); //构造矩阵
void pow(long long mat[][15]); //快速幂
void Printf(long long mat[][N]);

int main()
{
	long long mat[15][15];

	while (~scanf("%d %d", &n, &m))
	{
		int i, j, k;
		for (i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		if(m+n==0)
		{
			printf("0\n");
			continue;
		}
		a[0] = 23;
		a[n+1] = 3;
		makemat(mat);
		//Printf(mat);
		pow(mat);
	}
	return 0;
}
void makemat(long long mat[][N])
{
	int i, j;
	for (i = 0; i <= n+1; i++)
	{
		if (i == n+1)
		{
			for (j = 0; j < n+1; j++)
			{
				mat[i][j] = 0;
			}
				mat[i][j] = 1;
				break;
			}

		mat[i][0] = 10;
		for (j = 1; j <= n+1; j++)
		{
			if (j <= i )
			{
				mat[i][j] = 1;
				continue;
			}
			mat[i][j] = 0;
		}
		mat[i][n+1] = 1;
	}
}

void pow(long long mat[][15])
{
	long long b[15][15];
	//构造单位矩阵， 即 1 
	for (int i = 0; i <= n+1; i++)
	{
		for (int j = 0; j <= n+1; j++)
		{
			if (i == j)
			{
				b[i][j] = 1;
				continue;
			}
			b[i][j] = 0;
		}
	}

	while (m)
	{
		if (m & 1)
		{
			matric_mul(b, mat);
		}

		matric_mul(mat, mat);

		m >>= 1;
	}
	//Printf(b);
    long long sum = 0;
	for (int i = 0; i <= n+1; i++)
	{
		sum += a[i] * b[n][i]; //这样就行了， 只需b[n]行相乘就够了
	}

	printf("%d\n", sum%MOD);
}

void matric_mul(long long a[][N], long long b[][N])
{
	int i, j, k;
	long long tmp1[N][N] = {0};

	for (i = 0; i <= n+1; i++)
	{
		for (j = 0; j <= n+1 ; j++)
		{
			for (k = 0; k <= n+1; k++)
			{
				tmp1[i][j] = (tmp1[i][j] + b[i][k] * a[k][j]) % MOD;
			}
			//printf("tem[%d][%d] = %d\n", i, j, tmp1[i][j]);
		}
	}

	for (i = 0; i <= n+1; i++)
	{
		for (j = 0; j <= n+1; j++)
		{
			a[i][j] = tmp1[i][j];
		}
	}
}