Reziba has many magic gems. Each magic gem can be split into ??M normal gems. The amount of space each magic (and normal) gem takes is 11 unit. A normal gem cannot be split.
Reziba wants to choose a set of magic gems and split some of them, so the total space occupied by the resulting set of gems is ??N units. If a magic gem is chosen and split, it takes ??M units of space (since it is split into ??M gems); if a magic gem is not split, it takes 11 unit.
How many different configurations of the resulting set of gems can Reziba have, such that the total amount of space taken is ??N units? Print the answer modulo 10000000071000000007 (109+7109+7). Two configurations are considered different if the number of magic gems Reziba takes to form them differs, or the indices of gems Reziba has to split differ.
The input contains a single line consisting of 22 integers ??N and ??M (1≤??≤10181≤N≤1018, 2≤??≤1002≤M≤100).
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 10000000071000000007 (109+7109+7).
4 2
5
3 2
3
In the first example each magic gem can split into 22 normal gems, and we know that the total amount of gems are 44.
Let 11 denote a magic gem, and 00 denote a normal gem.
The total configurations you can have is:
Hence, answer is 55.
矩阵快速幂
1 #include <bits/stdc++.h> 2 #define M 101 3 using namespace std; 4 typedef long long ll; 5 ll n,m; 6 const int mod = 1e9+7; 7 struct Mat{ 8 ll a[M][M]; 9 Mat(){ 10 memset(a,0,sizeof(a)); 11 } 12 }; 13 Mat operator * (Mat A,Mat B){ 14 Mat ret; 15 for (int i = 0;i < m;++i){ 16 for (int j = 0;j < m;++j){ 17 ll tmp = 0; 18 for (int k = 0;k < m;++k){ 19 tmp = (tmp + A.a[i][k]*B.a[k][j]) % mod; 20 } 21 ret.a[i][j] = tmp; 22 } 23 } 24 return ret; 25 } 26 27 Mat power(Mat mat,ll n){ 28 Mat ret; 29 for (int i = 0;i < m;++i){ 30 ret.a[i][i] = 1; 31 } 32 while(n){ 33 if (n&1) ret = ret*mat; 34 n >>= 1; 35 mat = mat*mat; 36 } 37 return ret; 38 } 39 40 int main(){ 41 scanf("%lld%lld",&n,&m); 42 if (n < m) return 0 * puts("1"); 43 Mat mat; 44 mat.a[0][0] = mat.a[0][m-1] = 1; 45 for (int i = 1;i < m;++i) mat.a[i][i-1] = 1; 46 mat = power(mat,n-m+1); 47 ll ans = 0; 48 for (int i = 0;i < m;++i){ 49 ans = (ans + mat.a[0][i]) % mod; 50 } 51 printf("%lld\n",ans); 52 return 0; 53 }
[codefoeces] Educational Codeforces Round 60 D. Magic Gems
原文：https://www.cnblogs.com/mizersy/p/10402113.html