poj3233Matrix Power Series
也是矩阵经典题目 二分递归求解
a+a^2+a^3+..+a^(k/2)+a^(k/2+1)+...+a^k = a+a^2+..+a^k/2+a^k/2(a^1+a^2+..+a^k/2)(偶数）
a+a^2+a^3+..+a^(k/2)+a^(k/2+1)+...+a^k = a+a^2+..+a^k/2+a^k/2(a^1+a^2+..+a^k/2)+a^k。 奇数
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 1e9 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 struct Mat 18 { 19 int mat[31][31]; 20 }; 21 int n,mod; 22 Mat operator + (Mat a,Mat b) 23 { 24 Mat c; 25 int i,j; 26 for(i = 0 ; i < n ;i++) 27 for(j = 0 ;j < n ;j++) 28 { 29 if(a.mat[i][j]+b.mat[i][j]>mod) 30 c.mat[i][j] = (a.mat[i][j]+b.mat[i][j])%mod; 31 else 32 c.mat[i][j] = a.mat[i][j]+b.mat[i][j]; 33 } 34 return c; 35 } 36 Mat operator * (Mat a,Mat b) 37 { 38 Mat c; 39 memset(c.mat,0,sizeof(c.mat)); 40 int i,j,k; 41 for(k =0 ; k < n ; k++) 42 { 43 for(i = 0 ; i < n ;i++) 44 { 45 if(a.mat[i][k]==0) continue; 46 for(j = 0 ;j < n ;j++) 47 { 48 if(b.mat[k][j]==0) continue; 49 c.mat[i][j] = (c.mat[i][j]+a.mat[i][k]*b.mat[k][j])%mod; 50 } 51 } 52 } 53 return c; 54 } 55 Mat operator ^(Mat a,int k) 56 { 57 Mat c; 58 int i,j; 59 for(i =0 ; i < n ;i++) 60 for(j = 0; j < n ;j++) 61 c.mat[i][j] = (i==j); 62 for(; k ;k >>= 1) 63 { 64 if(k&1) c = c*a; 65 a = a*a; 66 } 67 return c; 68 } 69 Mat solve(Mat x,int k) 70 { 71 if(k==1) return x; 72 Mat c ; 73 c = x^k; 74 Mat a = solve(x,k/2); 75 Mat b = x^(k/2); 76 if(k&1) c = a+b*a+c; 77 else c = a+b*a; 78 return c; 79 } 80 int main() 81 { 82 int t; 83 int i,j; 84 while(scanf("%d%d%d",&n,&t,&mod)!=EOF) 85 { 86 Mat x; 87 for(i = 0 ; i < n ;i++) 88 for(j = 0; j < n ;j++) 89 scanf("%d",&x.mat[i][j]); 90 x = solve(x,t); 91 for(i =0 ; i < n ;i++) 92 { 93 for(j =0 ; j < n-1; j++) 94 printf("%d ",x.mat[i][j]%mod); 95 printf("%d\n",x.mat[i][n-1]%mod); 96 } 97 } 98 return 0; 99 }
