imut zcy,I promise You.

Sol：就是求第N项的斐波那契数。矩阵乘法+快速幂 #include #include #include #include using namespace std; #define LL long long struct Mat{ LL f[2][2]; }; LL MOD = 10000; Mat mul(Mat a,Mat b) { LL i,j,k

线性推，矩阵乘法+快速幂求通项。 传送门：点击打开链接 #include #include #include #include using namespace std; #define LL long long struct Mat{ LL f[2][2]; }; LL MOD; Mat mul(Mat a,Mat b) { LL i,j,k;

#include #include #include #include #include using namespace std; #define LL long long struct Mat{ LL f[3][3]; }; LL MOD = 10000007; int a[100010]; Mat mul(Mat a,Mat b) { LL i,j,k;