HDU - 5015 Matrix（矩阵快速幂）

233 Matrix Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 3402    Accepted Submission(s): 1903   Problem Description In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,m in the 233 matrix?   Input There are multiple test cases. Please process till EOF. For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).   Output For each case, output an,m mod 10000007.   Sample Input 1 1 1 2 2 0 0 3 7 23 47 16 扫描二维码关注公众号，回复： 2816747 查看本文章   Sample Output 234 2799 72937 Hint   Source 2014 ACM/ICPC Asia Regional Xi'an Online   Recommend hujie 这个题需要构建一个矩阵a：(大小是（n+2）*（n+2））（比如n=5） 10    0      0     0    0     0      1   10    1      0     0    0     0      1 10    1      1     0    0     0      1 10    1      1     1    0     0      1 10    1      1     1    1     0      1 10    1      1     1    1     1      1 0      0      0     0    0     0      1 然后a的（m-1）次幂与矩阵b相乘得矩阵c； 矩阵b大小为（（n+1）*1）；（比如n=5） 233 233+b[1][0] 233+b[1][0]+b[2][0] 233+b[1][0]+b[2][0]+b[3][0] 233+b[1][0]+b[2][0]+b[3][0]+b[4][0] 233+b[1][0]+b[2][0]+b[3][0]+b[4][0]+b[5][0] 3 得到的结果矩阵的c[n][0]就是所求的值 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <vector> #include <map> #include <string> #include <queue> #include <stack> #include <set> #include <list> #include <bitset> using namespace std; typedef long long ll; typedef unsigned long long ull; const int N = 15; //矩阵快速幂模板不变，只需要给N（矩阵大小），Mod（模大小），msize，nsize，（矩阵的行列大小）赋 //值，就可以用了 int Mod = 1e7+7; int n,m,nsize,msize; struct Mat{ ll mat[N][N]; }; Mat operator *(Mat a,Mat b) { Mat c; memset(c.mat,0,sizeof(c.mat)); int k,i,j; for(k = 0;k < msize;k++) for(i = 0;i < msize;i++) if(a.mat[i][k]) for(j = 0;j < nsize;j++) if(b.mat[k][j]) c.mat[i][j] = (c.mat[i][j] + a.mat[i][k]*b.mat[k][j]) % Mod; return c; } Mat operator ^ (Mat a,ll k) { Mat c; memset(c.mat,0,sizeof(c.mat)); for(int i = 0;i < msize;i++) c.mat[i][i] = 1; for(;k;k >>= 1){ if(k & 1) c = c*a; a = a*a; } return c; } int main() { while(cin >> n >> m){ msize = nsize = n+2; Mat a; memset(a.mat,0,sizeof(a.mat)); int i,j; for(i = 0;i <= n;i++) for(j = 0;j <= n+1;j++){ a.mat[i][0] = 10; a.mat[i][n+1] = 1; if(i>=j) a.mat[i][j] = 1; } a.mat[n+1][n+1] = 1; Mat b = a^(m-1); Mat c; c.mat[0][0] = 233; for(i = 1;i <= n;i++){ cin >> c.mat[i][0]; c.mat[i][0] += c.mat[i-1][0]; } c.mat[n+1][0] = 3; msize = n+2,nsize=1; Mat d = b*c; cout << d.mat[n][0]%Mod << endl; } return 0; }