POJ 3070 Fibonacci（矩阵快速幂）

求Fn (mod 10000)

Input

多组输入，每组用例占一行为一整数n，以n=-1结束输入

Output

对于每组用例，输出Fn（mod 10000）

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Solution

矩阵快速幂原题（注意矩阵初始化）

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>using namespace std;const int maxn = 2;struct Mat{    int mat[maxn][maxn]; //矩阵    int row, col;        //矩阵行列数};Mat mod_mul(Mat a, Mat b, int p) //矩阵乘法{    Mat ans;    ans.row = a.row;    ans.col = b.col;    memset(ans.mat, 0, sizeof(ans.mat));    for (int i = 0; i < ans.row; i++)        for (int j = 0; j < ans.col; j++)            for (int k = 0; k < a.col; k++)            {                ans.mat[i][j] += a.mat[i][k] * b.mat[k][j];                ans.mat[i][j] %= p;            }    return ans;}Mat mod_pow(Mat a, int k, int p) //矩阵快速幂{    Mat ans;    ans.row = a.row;    ans.col = a.col;    for (int i = 0; i < a.row; i++)        for (int j = 0; j < a.col; j++)            ans.mat[i][j] = (i == j);    while (k)    {        if (k & 1)            ans = mod_mul(ans, a, p);        a = mod_mul(a, a, p);        k >>= 1;    }    return ans;}int main(){    int n;    while (scanf("%d", &n) && n != -1)    {        Mat a;        a.mat[0][0] = a.mat[0][1] = a.mat[1][0] = 1;        a.mat[1][1] = 0;        a.col = a.row = 2;        Mat ans = mod_pow(a, n, 10000);        printf("%d\n", ans.mat[0][1]);    }    return 0;}