UVA11149 矩阵快速幂

首先我们来想一下计算A+A^2+A^3...+A^k。

如果A=2,k=6。那你怎么算

2+2²+2³+2⁴+2⁵+2⁶ = ？= (2+2²+2³)*(1+2³)

如果A=2,k=7。那你怎么算

2+2²+2³+2⁴+2⁵+2⁶+2⁷ = ？= (2+2²+2³)*(1+2³)+2⁷

^{so....同理：}

当k是偶数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))。

当k是奇数，A+A^2+A^3...+A^k=（E+A^(k/2)）*(A+A^2...+A^(k/2))+A^k。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cstdlib>
 using namespace std;
 #define MAXN 50
 int n,K;
 struct node
 {
     int mat[MAXN][MAXN];
 };
 node calcu(node x, node y)
 {
     node ret;
     memset(ret.mat,,sizeof(ret.mat));
     for (int i = ; i <= n ; i++)
         for (int j = ; j <= n ; j++)
     {
         for (int k = ; k <= n ; k++)
             ret.mat[i][j] = (ret.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % ;
     }
     return ret;
 }
 node add(node x,node y)
 {
     node ret;
     memset(ret.mat,,sizeof(ret.mat));
     for (int i = ; i <= n ; i++)
         for (int j = ; j <= n ; j++)
    {
            ret.mat[i][j] = x.mat[i][j] + y.mat[i][j];
            ret.mat[i][j] %= ;
    }
    return ret;
 }
 node pow_mat(node x,int cnt)
 {
     node ret;
     memset(ret.mat,,sizeof(ret.mat));
     for (int i = ; i < MAXN ; i++) ret.mat[i][i] = ;
     while (cnt)
     {
         if (cnt & ) ret = calcu(ret,x);
         x = calcu(x,x);
         cnt >>= ;
     }
     return ret;
 }
 node dfs(node cur, int k)
 {
     if (k == ) return cur;
     node res = dfs(cur,k / );
     node ans;
     ans = add(res,calcu(res,pow_mat(cur,k / )));
     if (k & ) ans = add(ans,pow_mat(cur,k));
     return ans;
 }
 int main()
 {
     while (scanf("%d%d",&n,&K) != EOF)
     {
         if (n == ) break;
         node ans;
         for (int i = ; i <= n ; i++)
                 for (int j = ; j <= n ; j++) {scanf("%d",&ans.mat[i][j]); ans.mat[i][j] %= ;}
         node ret = dfs(ans,K);
         for (int i = ; i <= n ; i++)
         {
                 printf("%d",ret.mat[i][]);
                 for (int j = ; j <= n ; j++)
                         printf(" %d",ret.mat[i][j]);
                 putchar('\n');
         }
         putchar('\n');
     }
     return ;
 }