CodeForces405B - Jzzhu and Sequences 矩阵快速幂

就是说：

f1 = x; f2 = y; 然后 f(i) = f(i-1) + f(i+1)

化简一下式子：

f(i+1) = f(i) - f(i-1)

用i 替换 i+1，则

f(i) = f(i-1) - f(i-2) 这样就差不多可斐波那契的矩阵快速幂了构造一个矩阵

$\begin{pmatrix} f1 & f2 \end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}=\begin{pmatrix} f2 & \quad f2-f1=f3 \end{pmatrix}$

就是很裸的矩阵快速幂了

注意取膜啊阿啊阿啊阿啊

#include <stdio.h>
typedef long long LL;
long long read(){long long x = 0, f = 1;char ch=getchar();while(ch < '0' || ch > '9'){if(ch=='-') f = -1; ch = getchar();}while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0'; ch = getchar();}return x * f;
}
const int maxn = 2;
const int mod = 1e9 + 7;
struct Matrix{long long int m[maxn][maxn];
}unit;
Matrix operator * (Matrix a, Matrix b){Matrix ret;LL x;int n = 2;for(int i = 0; i < n; ++i){for(int j = 0; j < n; ++j){x = 0;for(int k = 0; k < n; ++k){x += (((LL)a.m[i][k] * b.m[k][j]) + mod ) % mod;}ret.m[i][j] = (x + mod) % mod;}}return ret;
}
void init_unit(){// 单位矩阵for(int i = 0; i < maxn; ++i){unit.m[i][i] = 1;}
}
Matrix pow_mat(Matrix a, LL n){Matrix ans = unit;while (n) {if(n & 1){ans = ans * a;}a = a * a;n >>= 1;}return ans;
}
int main(){init_unit();Matrix ans, a;ans.m[0][0] = read();ans.m[0][1] = read();a.m[0][0] = 0; a.m[0][1] = -1;a.m[1][0] = 1; a.m[1][1] = 1;long long int n = read();a = pow_mat(a, n - 1);ans = ans * a;printf("%lld\n", (ans.m[0][0] + mod )% mod);return 0;
}

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CodeForces405B - Jzzhu and Sequences 矩阵快速幂

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