nyist 299 Matrix Power Series

在poj上是3233

nyist是299

这题就是给你一个矩阵A 求他的A + A^2 + A^2+.....A^n %m

就是一道矩阵的运算

首先看这个式子我们把它进行变化

F = A + A ^2 + A^3+... +A^n

n为偶数是：

　　 A+A^2+...A^n/2 + A^n/2(A+A^2 +...A^n/2)

n为基数时

　　A+A^2+..A^n/2+A^n^/2+1 + A^n/2+1(A+A^2+A^n/2)

计算A^n使用矩阵快速幂

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef struct numb
{
    long long  a[35][35];
    numb()
    {
        memset(a,0,sizeof(a));
    }
}numb;

numb x,z,y,tmp;
long long m,n,k;
void print(numb z)
{
    int i,j;
    for(i = 0 ;i < n;i++)
    {

        for(j = 0; j <n ;j++)
        printf("%lld ",z.a[i][j]) % m;
        printf("\n");
    }
}


void set_1(numb &tmp)//构建单位矩阵
{
    int i = 0;
    for(i = 0; i < n; i++)
        tmp.a[i][i] = 1;
}

/*void set_2(numb &tmp)//构建全1矩阵
{
    int i = 0;
    int j = 0;
    for(i = 0; i < n; i++)
    for(j = 0; j < n; j++)
        tmp.a[i][j]  = 1;
}
*/
numb mul(numb q,numb p)//矩阵A*B
{
    numb kk;

    int i = 0;
    int j = 0;
    int k = 0;
    long sum = 0;

    for(i = 0; i < n; i++)
    for(j = 0; j < n; j++)
        {
            sum = 0;
            for(k = 0; k < n; k++)
                sum += q.a[i][k]*p.a[k][j] % m;

            kk.a[i][j] = sum % m;
        }
        return kk;
}

numb add(numb x,numb y)//矩阵A+B
{
    numb tmp5;
    int i= 0;
    int j = 0;
    for(i = 0; i < n; i++)
    for(j = 0; j < n; j++)
     tmp5.a[i][j] = (x.a[i][j]+y.a[i][j]) % m;

     return tmp5;
}

numb kuaishu(numb x,int kk)//矩阵快速幂
{
    numb tmp2;
    set_1(tmp2);
    while(kk)
    {
        if(kk&1)
        {
            tmp2 = mul(tmp2,x);
        }
        kk >>=1;
        x = mul(x,x);
    }
    return tmp2;
}

numb sum(numb A,int kk)//求和
{
    numb tmp4;
    if(kk == 1)
        {
            tmp4 = A;
            return tmp4;
        }
    else
    {
        numb tmp3 = sum(A,kk/2);

        if(kk&1)
        {
            numb tmp2 = kuaishu(A,kk/2+1);

            tmp4 = add(add(tmp3,tmp2),mul(tmp2,tmp3));
        }
        else
        {
            numb tmp5 = kuaishu(A,kk/2);
            tmp4 = add(tmp3,mul(tmp5,tmp3));
        }

        return tmp4;
    }
}

int main()
{
    while(scanf("%lld%lld%lld",&n,&k,&m) != EOF)
    {
        int i = 0;
        int j = 0;
        for(i = 0; i < n; i++)
        for(j = 0; j < n; j++)
            scanf("%lld",&x.a[i][j]);
       z = sum(x,k);
        print(z);
    }
    return 0;
}