UVA - 12470 Tribonacci 矩阵快速幂

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题目大意：fibonacci的升级版，规则是f(n) = f(n-1) + f(n-2) + f(n-3)

解题思路：水题

#include<cstdio>typedef long long ll;const int N = 3;const ll mod = 1e9 + 9;struct Matrix{    ll mat[N][N];}A, B, tmp;ll 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;            if(i == j)                 B.mat[i][j] = 1;        }    A.mat[0][0] = A.mat[1][0] = A.mat[2][0] = A.mat[0][1] = A.mat[1][2] = 1;}Matrix matMul(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] = (tmp.mat[i][j] + (x.mat[i][k] * y.mat[k][j]) % mod ) % mod;        }    return tmp;}void solve() {    while(n) {        if(n & 1)             B = matMul(B,A);        A = matMul(A,A);        n >>= 1;    }}int main() {    while(scanf("%lld", &n) != EOF && n) {        if(n == 3 || n == 1 || n == 2) {            printf("%lld\n", n - 1);            continue;        }        n -= 3;        init();        solve();        ll ans = 0;        ans = (ans + 2 * B.mat[0][0]) % mod;        ans = (ans + B.mat[1][0]) % mod;        printf("%lld\n", ans);    }    return 0;}

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