Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems（动态规划+矩阵快速幂）

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems

Time Limit: 3000 mSec

Problem Description

Input

The input contains a single line consisting of 2 integers N and M (1≤N≤10^18, 2≤M≤100).

Output

Print one integer, the total number of configurations of the resulting set of gems, given that the total amount of space taken is N units. Print the answer modulo 1000000007.

Sample Input

4 2

Sample Output

5

题解：很简单的dp，考虑第一个数分裂不分裂，dp[n] = dp[n - m] + dp[n - 1]，其中dp[n]就是站n个单位空间的方法数，由于N比较大，所以很明显要用矩阵快速幂，写这篇题解同时提醒自己快速幂之前一定要判断指数是否非负。

 #include <bits/stdc++.h>

 using namespace std;

 #define REP(i, n) for (int i = 1; i <= (n); i++)
 #define sqr(x) ((x) * (x))

 const int maxn =  + ;
 const int maxm =  + ;
 const int maxs =  + ;

 typedef long long LL;
 typedef pair<int, int> pii;
 typedef pair<double, double> pdd;

 const LL unit = 1LL;
 const int INF = 0x3f3f3f3f;
 const double eps = 1e-;
 const double inf = 1e15;
 const double pi = acos(-1.0);
 const int SIZE =  + ;
 const LL MOD = ;

 struct Matrix
 {
     int r, c;
     LL mat[SIZE][SIZE];
     Matrix() {}
     Matrix(int _r, int _c)
     {
         r = _r;
         c = _c;
         memset(mat, , sizeof(mat));
     }
 };

 Matrix operator*(Matrix a, Matrix b)
 {
     Matrix s(a.r, b.c);
     for (int i = ; i < a.r; i++)
     {
         for (int k = ; k < a.c; k++)
         { //j, k交换顺序运行更快
             for (int j = ; j < b.c; j++)
             {
                 s.mat[i][j] = (s.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % MOD;
             }
         }
     }
     return s;
 }

 Matrix pow_mod(Matrix a, LL n)
 {
     Matrix ret(a.r, a.c);
     for (int i = ; i < a.r; i++)
     {
         ret.mat[i][i] = ; //单位矩阵
     }
     Matrix tmp(a);
     while (n)
     {
         if (n & )
             ret = ret * tmp;
         tmp = tmp * tmp;
         n >>= ;
     }
     return ret;
 }

 LL n, m;

 int main()
 {
     ios::sync_with_stdio(false);
     cin.tie();
     //freopen("input.txt", "r", stdin);
     //freopen("output.txt", "w", stdout);
     cin >> n >> m;
     if (n < m)
     {
         cout <<  << endl;
         return ;
     }
     Matrix ori = Matrix(m, );
     for (int i = ; i < m; i++)
     {
         ori.mat[i][] = unit;
     }
     Matrix A = Matrix(m, m);
     A.mat[][] = unit;
     A.mat[][m - ] = unit;
     for (int i = ; i < m; i++)
     {
         A.mat[i][i - ] = unit;
     }
     Matrix ans = pow_mod(A, n - m + );
     ori = ans * ori;
     cout << ori.mat[][] << endl;
     return ;
 }