poj 3233 Matrix Power Series 矩阵快速幂

点击打开链接 题目链接

Matrix Power Series

Time Limit: 3000MS		Memory Limit: 131072K
Total Submissions: 14708		Accepted: 6313

Description

Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.

Input

The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative
integers below 32,768, giving A’s elements in row-major order.

Output

Output the elements of S modulo m in the same way as A is given.

Sample Input

2 2 4
0 1
1 1

Sample Output

1 2
2 3

给出n*n矩阵 ，k，m

求A + A2 + A3 +
… + Ak.各项%m的值

若k为偶数 S(K) =(1 + A^k/2) * S(K/2)。

若K为奇数 S(K)
=A + (A + A^(k/2+1)) * S(k/2)。

贴代码

#include<stdio.h>
#include<string.h>
int n,m,k;
struct juzheng
{
	int v[30][30];
};
juzheng a;
juzheng danwei;
juzheng mot(juzheng a,juzheng b)
{
	juzheng c;
	int i,j,k;
	for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
		{
			c.v[i][j]=0;
			for(k=0;k<n;k++)
			{
				c.v[i][j]+=a.v[i][k]*b.v[k][j]%m;
			}
			c.v[i][j]%=m;
		}
	}
	return c;
}
juzheng add(juzheng a,juzheng b)
{
	juzheng c;
	int i,j;
	for(i=0;i<n;i++)
	{
		for(j=0;j<n;j++)
		{
			c.v[i][j]=(a.v[i][j]+b.v[i][j])%m;
		}
	}
	return c;
}
juzheng pow(juzheng a,int t)
{
	juzheng c;
	int i,j;
	for(i=0;i<n;i++)
    {
        for(j=0;j<n;j++)
        {
            if(i==j)
                c.v[i][j]=1;
            else c.v[i][j]=0;
        }
    }
	while(t)
	{
		if(t%2)
			c=mot(c,a);
	    a=mot(a,a);
	    t/=2;
	}
	return c;
}
juzheng sum(juzheng a,int t)
{
	juzheng b,c;
	int i,j;
	if(t==1)
	{
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
				b.v[i][j]=a.v[i][j];
			}
		}
		return b;
	}
	else if(t%2)
	{
		c=pow(a,t/2+1);
		c=add(c,a);
		b=sum(a,t/2);
		return add(mot(c,b),a);
	}
	else
	{
		c=pow(a,t/2);
		c=add(c,danwei);
		b=sum(a,t/2);
		return mot(c,b);
	}
}
int main()
{
	int i,j;
	while(scanf("%d%d%d",&n,&k,&m)!=EOF)
	{
		juzheng b;
		for(i=0;i<n;i++)
		{
		    for(j=0;j<n;j++)
		    {
		    	scanf("%d",&a.v[i][j]);
		    }
		}
		for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(i==j)
                danwei.v[i][j]=1;
                else danwei.v[i][j]=0;
            }
        }
		b=sum(a,k);
		for(i=0;i<n;i++)
		{
			for(j=0;j<n;j++)
			{
                if(j<n-1)
				printf("%d ",b.v[i][j]);
			    else printf("%d",b.v[i][j]);
			}
			printf("\n");
		}
	}
	return 0;
}