Classic matrix Question 7: Warcraft III daemon (recursive matrix acceleration)

Https://www.vijos.org/p/1067

It is easy to introduce the recursive formula F [N] = f [n-1] + F [N-2] +... + F [n-k].

Methods of constructing a matrix: To construct a K * k matrix, where the upper right corner of the (k-1) * (k-1) matrix is the unit matrix, each number in the k-th row corresponds to the coefficients of F [n-1], F [N-2], and f [n-k] respectively. Then, a K * 1 matrix is constructed. Its line I represents f [I], which is obtained through direct recursion. Set ans [] [] to the n-k power of the first matrix to multiply the second matrix. f [N] Is ans [k] [1].

Note: Use _ int64

#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 long long#define _LL __int64#define eps 1e-12#define PI acos(-1.0)#define C 240#define S 20using namespace std;const int maxn = 15;const int mod = 7777777;int k;struct matrix{_LL mat[maxn][maxn];void init(){memset(mat,0,sizeof(mat));for(int i = 1; i <= maxn; i++)mat[i][i] = 1;}}a,b;matrix mul(matrix a, matrix b){matrix ans;memset(ans.mat,0,sizeof(ans.mat));for(int i = 1; i <= k; i++){for(int g = 1; g <= k; g++){if(a.mat[i][g] == 0) continue;for(int j = 1; j <= k; j++){ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][g] * b.mat[g][j])%mod;}}}return ans;}matrix pow(matrix a, int n){matrix ans;ans.init();while(n){if(n&1)ans = mul(ans,a);a = mul(a,a);n >>= 1;}return ans;}int main(){int n;while(~scanf("%d %d",&k,&n)){memset(a.mat,0,sizeof(a.mat));for(int i = 1; i <= k-1; i++)a.mat[i][i+1] = 1;for(int i = 1; i <= k; i++)a.mat[k][i] = 1;matrix ans = pow(a,n-k);memset(b.mat,0,sizeof(b.mat));b.mat[0][1] = 1;for(int i = 1; i <= k; i++){for(int j = 0; j < i; j++)b.mat[i][1] += b.mat[j][1];}ans = mul(ans,b);printf("%I64d\n",ans.mat[k][1]);}return 0;}