題意:公式f(n) = f(n - 1) + f(n - 2) + f(n - 3),給出n,f(1) = 0,f(2) = 1, f(3) = 2,要求得出f(n)。
題解:普通的矩陣快速冪模板題。
#include
#include
const int MOD = 1000000009;
struct Mat {
long long g[3][3];
}ori, res;
long long n;
Mat multiply(Mat x, Mat y) {
Mat temp;
for (int i = 0; i < 3; i++)
for (int j = 0; j < 3; j++) {
temp.g[i][j] = 0;
for (int k = 0; k < 3; k++)
temp.g[i][j] = (temp.g[i][j] + x.g[i][k] * y.g[k][j]) % MOD;
}
return temp;
}
void calc(long long n) {
while (n) {
if (n & 1)
ori = multiply(ori, res);
n >>= 1;
res = multiply(res, res);
}
}
int main() {
while (scanf("%lld", &n) && n) {
if (n == 1 || n == 2 || n == 3) {
printf("%lld\n", n - 1);
continue;
}
memset(ori.g, 0, sizeof(ori));
memset(res.g, 0, sizeof(res));
ori.g[0][0] = 2;
ori.g[0][1] = 1;
ori.g[0][2] = 0;
res.g[0][0] = res.g[1][0] = res.g[2][0] = 1;
res.g[0][1] = res.g[1][2] = 1;
calc(n - 3);
printf("%lld\n", ori.g[0][0]);
}
return 0;
}