Poj 3233 Matrix Power Series（矩阵二分快速幂）

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<iostream>
  4 #include<algorithm>
  5 #include<deque>
  6 #include<cmath>
  7 using namespace std;
  8 typedef __int64 INT;
  9 const int N = 31;
 10 int m;
 11 
 12 struct node
 13 {
 14     INT t[N][N];
 15     int n;
 16     friend node operator * (node a,node b)
 17     {
 18         node B = a;
 19         for(int i = 0;i < a.n;++i)
 20         for(int j = 0;j < a.n;++j)
 21         {
 22             int tot = 0;
 23             for(int k = 0;k < a.n;++k)
 24             tot += ((a.t[i][k] * b.t[k][j]) % m);
 25             B.t[i][j] = tot % m;
 26         }
 27         return B;
 28     }
 29     friend node operator + (node a,node b)
 30     {
 31         node B;
 32         B.n = a.n;
 33         for(int i = 0;i < a.n;++i)
 34         for(int j = 0;j < a.n;++j)
 35         B.t[i][j] = (a.t[i][j] + b.t[i][j]) % m;
 36         return B;
 37     }
 38 };
 39 node A;
 40 void print(node t)
 41 {
 42     for(int i = 0;i < t.n;++i)
 43     for(int j = 0;j < t.n;++j)
 44     printf(j == t.n-1? "%d\n":"%d ",t.t[i][j]);
 45 }
 46 node Pw(node tt,int n)   //二分矩阵快速幂求A ^ n 
 47 {
 48     if(n == 1) return A;
 49     node res;
 50     res.n = tt.n;
 51     memset(res.t,0,sizeof(res.t));
 52     for(int i = 0;i < res.n;++i)
 53     res.t[i][i] = 1;
 54     while(n)
 55     {
 56         if(n & 1)
 57         res = res * tt;
 58         n >>= 1;
 59         tt = tt * tt;
 60     }
 61     return res;
 62 }    
 63 
 64 node get_ans(int n)    //求A^1 到 A ^ n的和 
 65 {
 66     if(n == 1)
 67     return A;
 68     if(n & 1)
 69     {
 70         node temp1;
 71         temp1.n = A.n;
 72         temp1 = Pw(A,n);
 73         node temp2;
 74         temp2.n = 2;
 75         temp2 = get_ans(n-1);
 76         temp1 = temp1 + temp2;
 77         return temp1;
 78     }
 79     else
 80     {
 81         node temp1;
 82         temp1.n = A.n;
 83         temp1  = Pw(A,n / 2);
 84         node temp2 = get_ans(n / 2);
 85         temp2.n = A.n;
 86         temp1 = temp1 * temp2;
 87         temp1 = temp1 + temp2;
 88         return temp1;
 89     }
 90 }
 91 
 92 int main()
 93 {
 94     
 95     int n,k;
 96     while(scanf("%d%d%d",&n,&k,&m)!=EOF)
 97     {
 98         A.n = n;
 99         for(int i = 0;i < n;++i)
100         for(int j = 0;j < n;++j)
101         scanf("%d",&A.t[i][j]);
102         node ans = get_ans(k);
103         print(ans);
104     }
105     return 0;
106 }