矩阵快速幂

问 题 简 述:

求Fibonacci数列第n项 F[n] % 10000。其中: n ≤ 10086110    询问次数 Q ≤ 10  

Hint:
Fibonacci数列定义：
F[0] = 0
F[1] = 1
F[n] = F[n − 1] + F[n − 2] （ n ≥ 2）

#include <cstdio>int A[3], B[3], temp[3], n, q;void mul (int* X, int* Y ) {temp[0]  =  X[0] * Y[0] + X[1] * Y[1]  % 10000;temp[1]  =  X[0] * Y[1] + X[1] * Y[2]  % 10000;temp[2]  =  X[1] * Y[1] + X[2] * Y[2]  % 10000;for ( int i =  0; i < 3; ++i)  X[i]  = temp[i];}void pow()   {for( ;n; n >>=  1, mul (A, A))  if(n&1==1)  mul (B,A);}int main()   {scanf("%d", &q);while(q-- ) {scanf ("%d", &n) ;A[0]=A[1]=B[0]=B[2]=1, A[2]=B[1]=0;pow();printf ("%d\n", B[1]) ;}return 0;}                /*输入示例 209输出示例034*/