http://poj.org/problem?id=3233

Matrix Power Series
Time Limit: 3000MS Memory Limit: 131072K
Total Submissions: 8740 Accepted: 3739

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

代码：

#include<iostream>
#include<cstdio>
#include<string.h>
#define N 35
using namespace std;
typedef struct{  int num[N][N];
        }Node;
Node p,q;
int n,m,k;
Node add(Node a,Node b)
{  Node c;
    memset(c.num,0,sizeof(c.num));
    for(int i=0;i<n;++i)
      for(int j=0;j<n;++j)
       c.num[i][j]= (a.num[i][j]+b.num[i][j])%m;
       return c;
}
Node mul(Node a,Node b)
{    Node c;
  memset(c.num,0,sizeof(c.num));
     for(int i=0;i<n;++i)
       for(int j=0;j<n;++j)
       {   for(int t=0;t<n;++t)
           c.num[i][j]= (c.num[i][j]+a.num[i][t]*b.num[t][j])%m;
       }
       return c;
}
Node ceil(int k)
{    Node a=p,b=q;
     while(k!=1)
     {  if(1&k) a=mul(a,b),k--;
        else{k=k/2;
             b=mul(b,b);
             }
     
     }
     a=mul(a,b);
     return a; 
}
/* 不知道这里为什么错，都是求矩阵的k次方为什么这个就不行呢，，纠结，，，不过在我的努力下我找到了。。。<img alt="微笑" src="http://static.blog.csdn.net/xheditor/xheditor_emot/default/smile.gif">
Node ceil(int k)
{    Node a=p,b=q;
     while(k)
     {
     if(1&k)  a=mul(a,b);
        b=mul(b,b);
        k=k>>1;
     }
     return a;  
}
*/
Node get_num(int k)
{ if(k==1) return q;
   Node a,b;
   a=get_num(k/2);
   if(1&k)
   {  b=ceil(k/2+1);
      a=add(a,mul(a,b));
      a=add(a,b);
   }
   else{  b=ceil(k/2);
          a=add(a,mul(b,a));
        }
        return a;
}
void print_f(Node a)
{   for(int i=0;i<n;++i)
       {for(int j=0;j<n;++j)
         printf("%d ",a.num[i][j]%m);
          printf("\n");
       }
}
int main()
{   while(~scanf("%d%d%d",&n,&k,&m))
    { for(int i=0;i<n;++i)
       for(int j=0;j<n;++j)
         { p.num[i][j]=(i==j);
            scanf("%d",&q.num[i][j]);
             q.num[i][j]%=m;
         }
         Node ans=get_num(k);
         print_f(ans);
         
         }
         return 0;
    }
        
        

今天做了两个关于矩阵的题，主要思想都是二分，，，看来二分很重要啊，，，还有就是见证了矩阵的强大，，，(*^__^*) 嘻嘻……，，

杯具的是我把2的n次方写成n>>2,杯具啊，，，，几乎杯具了一下午，，，，，

2的n次方n>>1,

代码：

#include<iostream>
#include<cstdio>
#include<string.h>
#define N 35
using namespace std;
typedef struct{  int num[N][N];
        }Node;
Node p,q;
int n,m,k;
Node add(Node a,Node b)
{  Node c;
    memset(c.num,0,sizeof(c.num));
    for(int i=0;i<n;++i)
      for(int j=0;j<n;++j)
       c.num[i][j]= (a.num[i][j]+b.num[i][j])%m;
       return c;
}
Node mul(Node a,Node b)
{    Node c;
  memset(c.num,0,sizeof(c.num));
     for(int i=0;i<n;++i)
       for(int j=0;j<n;++j)
       {   for(int t=0;t<n;++t)
           c.num[i][j]= (c.num[i][j]+a.num[i][t]*b.num[t][j])%m;
       }
       return c;
}
Node ceil(int k)
{    Node a=p,b=q;
     while(k)
     {
     if(1&k)  a=mul(a,b);
        b=mul(b,b);
        k=k>>1;
     }
     return a;  
}
Node get_num(int k)
{ if(k==1) return q;
   Node a,b;
   a=get_num(k/2);
   if(1&k)
   {  b=ceil(k/2+1);
      a=add(a,mul(a,b));
      a=add(a,b);
   }
   else{  b=ceil(k/2);
          a=add(a,mul(b,a));
        }
        return a;
}
void print_f(Node a)
{   for(int i=0;i<n;++i)
       {for(int j=0;j<n;++j)
         printf("%d ",a.num[i][j]%m);
          printf("\n");
       }
}
int main()
{   while(~scanf("%d%d%d",&n,&k,&m))
    { for(int i=0;i<n;++i)
       for(int j=0;j<n;++j)
         { p.num[i][j]=(i==j);
            scanf("%d",&q.num[i][j]);
             q.num[i][j]%=m;
         }
         Node ans=get_num(k);
         print_f(ans);
         
         }
         return 0;
    }