poj3233

题意：给出矩阵A，求S = A + A² + A³ + … + A^k

分析：把问题转化以加速，令

B = A  I

      0  I

则B^(k + 1) = A^(k + 1)      I + A + A² + A³ + … + A^k

                            0                          I

用二分法求B^(k + 1)

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
usingnamespace std;

#define maxn 61

struct Matrix
{
    int ma[maxn][maxn];
} b, ans;

int n, k, m;

void init()
{
    memset(b.ma, 0, sizeof(b.ma));
    memset(ans.ma, 0, sizeof(ans.ma));
    scanf("%d%d%d", &n, &k, &m);
    for (int i =0; i < n; i++)
        for (int j =0; j < n; j++)
        {
            scanf("%d", &b.ma[i][j]);
            b.ma[i][j] %= m;
            ans.ma[i][j] = b.ma[i][j];
        }
    for (int i =0; i < n; i++)
    {
        b.ma[i][n + i] =1;
        b.ma[n + i][n + i] =1;
        ans.ma[i][n + i] =1;
        ans.ma[n + i][n + i] =1;
    }
}

void mul(Matrix &a, Matrix &b)
{
    Matrix x;

    for (int i =0; i < n *2; i++)
        for (int j =0; j < n *2; j++)
            x.ma[i][j] = a.ma[i][j];
    for (int i =0; i < n *2; i++)
        for (int j =0; j < n *2; j++)
        {
            int sum =0;
            for (int k =0; k < n *2; k++)
                sum += x.ma[i][k] * b.ma[k][j];
            a.ma[i][j] = sum % m;
        }
}

void sqr(Matrix &a)
{
    Matrix x;

    for (int i =0; i < n *2; i++)
        for (int j =0; j < n *2; j++)
            x.ma[i][j] = a.ma[i][j];
    mul(a, x);
}

void work(Matrix &a, int num)
{
    if (num ==1)
        return;
    work(a, num /2);
    sqr(a);
    if (num %2==1)
    {
        mul(a, b);
    }
}

int main()
{
    freopen("D:\\t.txt", "r", stdin);
    init();
    work(ans, k +1);
    for (int i =0; i < n; i++)
    {
        if (ans.ma[i][n + i] ==0)
            ans.ma[i][n + i] = m;
        ans.ma[i][n + i] -=1;
    }
    for (int i =0; i < n; i++)
    {
        printf("%d", ans.ma[i][n]);
        for (int j =1; j < n; j++)
            printf(" %d", ans.ma[i][j + n]);
        printf("\n");
    }
    return0;
}