hihoCoder 1143 矩阵快速幂

传送门： HihoCoder 1143

题解

递推：

f
= f[n - 1] + f[n - 2];

过程： 放2* n个骨牌时（n >= 3）最后一列竖直放方案数为f[n - 2]最后两列放两个横放， 为f[n - 2]得证

构造：

(1110) * (f[n−1]f[n−2]) = (f[n]f[n−1])

(1110) ^ (n - 2) * (f[2]f[1]) = (f[n]f[n−1])(n >= 3)

注意一点， 矩阵乘法模数为19991997， 所以int可能会溢出， 应该用long long

code：

/*
adrui's submission
Language : C++
Result : Accepted
Love : ll
Favorite : Dragon Balls

Standing in the Hall of Fame
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;

#define debug 0
#define LL long long
#define mod 19999997
#define M(a, b) memset(a, b, sizeof(a))

int n;

struct Matrix {
LL mat[2][2];
void init() {
M(mat, 0);
for (int i = 0; i < 2; i++)
mat[i][i] = 1;
}
void Debug() {
for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
printf("%I64d%s", this->mat[i][j], j == 1 ? "\n" : " ");
}
};

Matrix operator * (Matrix a, Matrix t) {                //矩阵乘法
Matrix c;
M(c.mat, 0);

for (int i = 0; i < 2; i++)
for (int j = 0; j < 2; j++)
{
for (int k = 0; k < 2; k++)
c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * t.mat[k][j]) % mod;
}

return c;
}
Matrix operator ^ (Matrix tmp, int b) {                    //快速幂
Matrix res;
res.init();

while (b) {
if (b & 1) res = res * tmp;
tmp = tmp * tmp;
b >>= 1;
}

return res;
}
int main() {
#if debug
freopen("in.txt", "r", stdin);
#endif //debug

while (~scanf("%d", &n)) {
if (n < 3)
printf("%d\n", n);
else {
Matrix tmp;
M(tmp.mat, 0);           //构造底数矩阵
tmp.mat[0][0] = tmp.mat[0][1] = 1;
tmp.mat[1][0] = 1;

//tmp.Debug();
Matrix res = tmp ^ (n - 2);         //快速幂
//res.Debug();

int ans = (2 * res.mat[0][0] + res.mat[0][1]) % mod;

printf("%d\n", ans);
}
}
return 0;
}