hdu 5015 233 Matrix(构造矩阵)

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http://acm.hdu.edu.cn/showproblem.php?pid=5015

因为是个二维的递推式，当时没有想到可以这样构造矩阵。从列上看，当前这一列都是由前一列递推得到。根据这一点来构造矩阵。令b[i]代表第i列，是一个(n+2)*1的矩阵,即b[1] = [1,233......]，之所以在加了两行，是要从前一个矩阵b[i-1]得到b[i]中的第二个数2333...，再构造一个转换矩阵a,它是一个（n+2）*（n+2）的矩阵，那么a^(m-1) * b就是第m列。

/*a矩阵：1 0 0 0 0...3 10 0 0 0...3 10 1 0 0...3 10 1 1 0...3 10 1 1 1.....b矩阵：第1列*/#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>#define LL __int64//#define LL long long#define eps 1e-9#define PI acos(-1.0)using namespace std;const int INF = 0x3f3f3f3f;const int mod = 10000007;struct matrix{LL mat[15][15];void init(){memset(mat,0,sizeof(mat));for(int i = 0; i < 15; i++){mat[i][i] = 1;}}}a,b,res;int n,m;matrix mul(matrix a, matrix b){matrix ans;memset(ans.mat,0,sizeof(ans.mat));for(int i = 0; i < n+2; i++){for(int k = 0; k < n+2; k++){if(a.mat[i][k] == 0)continue;for(int j = 0; j < n+2; j++){ans.mat[i][j] += (a.mat[i][k] * b.mat[k][j])%mod;ans.mat[i][j] %= mod;}}}return ans;}matrix pow(matrix a, int n){matrix ans;ans.init();while(n){if(n&1)ans = mul(ans,a);n >>= 1;a = mul(a,a);}return ans;}int main(){int x;while(~scanf("%d %d",&n,&m)){memset(b.mat,0,sizeof(b.mat));b.mat[0][0] = 1;b.mat[1][0] = 233;for(int i = 2; i < n+2; i++){scanf("%d",&x);b.mat[i][0] = (b.mat[i-1][0] + x%mod)%mod;}memset(a.mat,0,sizeof(a.mat));a.mat[0][0] = 1;for(int i = 1; i < n+2; i++){a.mat[i][0] = 3;a.mat[i][1] = 10;for(int j = 2; j <= i; j++)a.mat[i][j] = 1;}res = pow(a,m-1);LL anw = 0;for(int i = 0; i < n+2; i++){anw += (res.mat[n+1][i] * b.mat[i][0])%mod;anw %= mod;}printf("%I64d\n",anw);}return 0;}

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