hdu1757---A Simple Math Problem(矩阵)
2015-03-11 20:33
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problem Description
Lele now is thinking about a simple function f(x).
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
Input
The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
Output
For each case, output f(k) % m in one line.
Sample Input
10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0
Sample Output
45 104
Author
linle
Source
2007省赛集训队练习赛(6)_linle专场
Recommend
lcy | We have carefully selected several similar problems for you: 1588 3117 2276 2256 2254
很明显的矩阵递推题
Lele now is thinking about a simple function f(x).
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
Input
The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
Output
For each case, output f(k) % m in one line.
Sample Input
10 9999 1 1 1 1 1 1 1 1 1 1 20 500 1 0 1 0 1 0 1 0 1 0
Sample Output
45 104
Author
linle
Source
2007省赛集训队练习赛(6)_linle专场
Recommend
lcy | We have carefully selected several similar problems for you: 1588 3117 2276 2256 2254
很明显的矩阵递推题
/************************************************************************* > File Name: hdu1757.cpp > Author: ALex > Mail: zchao1995@gmail.com > Created Time: 2015年03月11日 星期三 19时57分20秒 ************************************************************************/ #include <map> #include <set> #include <queue> #include <stack> #include <vector> #include <cmath> #include <cstdio> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; const double pi = acos(-1); const int inf = 0x3f3f3f3f; const double eps = 1e-15; typedef long long LL; typedef pair <int, int> PLL; int mod; class MARTIX { public: int mat[105][105]; MARTIX(); MARTIX operator + (const MARTIX &b)const; MARTIX operator * (const MARTIX &b)const; MARTIX & operator = (const MARTIX &b); }; MARTIX :: MARTIX() { memset (mat, 0, sizeof(mat)); } MARTIX MARTIX :: operator + (const MARTIX &b)const { MARTIX ret; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 10; ++j) { ret.mat[i][j] = mat[i][j] + b.mat[i][j]; ret.mat[i][j] %= mod; } } return ret; } MARTIX MARTIX :: operator * (const MARTIX &b)const { MARTIX ret; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 10; ++j) { ret.mat[i][j] = 0; for (int k = 0; k < 10; ++k) { ret.mat[i][j] += mat[i][k] * b.mat[k][j]; ret.mat[i][j] %= mod; } } } return ret; } MARTIX & MARTIX :: operator = (const MARTIX &b) { for (int i = 0; i < 10; ++i) { for (int j = 0; j < 10; ++j) { this -> mat[i][j] = b.mat[i][j]; } } return *this; } MARTIX fastpow(MARTIX A, LL n) { MARTIX ans; for (int i = 0; i < 10; ++i) { for (int j = 0; j < 10; ++j) { ans.mat[i][j] = (i == j); } } while (n) { if (n & 1) { ans = ans * A; } n >>= 1; A = A * A; } return ans; } void Debug(MARTIX A) { for (int i = 0; i < 10; ++i) { for (int j = 0; j < 10; ++j) { printf("%d ", A.mat[i][j]); } printf("\n"); } } int main () { LL k; while (~scanf("%lld%d", &k, &mod)) { if (k < 10) { printf("%lld\n", k); continue; } MARTIX A; for (int i = 0; i < 10; ++i) { scanf("%d", &A.mat[i][0]); } for (int i = 0; i < 9; ++i) { A.mat[i][i + 1] = 1; } MARTIX ans = fastpow(A, k - 9); MARTIX F; for (int i = 0; i < 10; ++i) { F.mat[0][i] = 10 - i - 1; } // Debug(F); ans = F * ans; printf("%d\n", ans.mat[0][0]); } return 0; }
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