FZU-1683 commemorating the SlingShot matrix power

FZU-1683 commemorating the SlingShot matrix power

Known F (n) = 3 * F (n-1) + 2 * F (n-2) + 7 * F (n-3), n> = 3, where F (0) = 1, F (1) = 3, F (2) = 5, for each given n, the output F (0) + F (1) + ...... + F (n) mod 2009.

Solution: Borrow others' pictures

This question is similar to HDU-1757 A Simple Math Problem, but this question has many similarities. In fact, the idea is similar. <喎?http: www.bkjia.com kf ware vc " target="_blank" class="keylink"> VcD4NCjxwcmUgY2xhc3M9 "brush: java;"> #include #include #define mod 2009const int N = 4;typedef long long ll;struct Matrix{ ll mat[N][N];}a, b, tmp;int n;void init() { for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) a.mat[i][j] = b.mat[i][j] = 0; for(int i = 0; i < N; i++) b.mat[i][i] = 1; a.mat[0][0] = a.mat[0][1] = a.mat[2][1] = a.mat[3][2] = 1; a.mat[1][1] = 3; a.mat[1][2] = 2; a.mat[1][3] = 7;}Matrix matrixMul(Matrix x, Matrix y) { for(int i = 0; i < N; i++) for(int j = 0; j < N; j++) { tmp.mat[i][j] = 0; for(int k = 0; k < N; k++) tmp.mat[i][j] += (x.mat[i][k] * y.mat[k][j]) % mod; } return tmp;}void solve() { while(n) { if(n & 1) b = matrixMul(b,a); a = matrixMul(a,a); n >>= 1; }}int main() { int test, cas = 1; scanf("%d", &test); while(test--) { scanf("%d", &n); init(); if(n <= 2) { switch(n) { case 0:printf("Case %d: 1\n", cas++);break; case 1:printf("Case %d: 4\n");break; case 2:printf("Case %d: 9\n");break; } continue; } n -= 1; solve(); printf("Case %d: %lld\n",cas++, (b.mat[0][0] * 4 + b.mat[0][1] * 5 + b.mat[0][2] * 3 + b.mat[0][3] * 1) % mod); } return 0;}