poj3070Fibonacci（矩阵快速幂模板）

代码一：

#include<iostream>
#include<cstdio>
using namespace std;
const int MOD = 10000;
int fast_mod(int n)
{
    int base[2][2] = {1, 1, 1, 0};
    int ans[2][2] = {1, 0, 0, 1};
    int tmp[2][2];
    while(n)
    {
        if(n & 1)
        {
            for(int i = 0; i < 2; i++)
                for(int j = 0; j < 2; j++)
                    tmp[i][j] = ans[i][j];
            ans[0][0] = ans[1][1] = ans[0][1] = ans[1][0] = 0;
            for(int i = 0; i < 2; i++)
            {
                for(int j = 0; j < 2; j++)
                {
                    for(int k = 0; k < 2; k++)
                       ans[i][j] = (ans[i][j] + base[i][k] * tmp[k][j]) % MOD;
                }
            }
        }
        for(int i = 0; i < 2; i++)
            for(int j = 0; j < 2; j++)
                tmp[i][j] = base[i][j];
        base[0][0] = base[0][1] = base[1][0] = base[1][1] = 0;
        for(int i = 0; i < 2; i++)
        {
            for(int j = 0; j < 2; j++)
            {
                for(int k = 0; k < 2; k++)
                    base[i][j] = (base[i][j] + tmp[i][k] * tmp[k][j]) % MOD;
            }
        }
        n >>= 1;
    }
    return ans[0][1];
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF && n != -1)
    {
        printf("%d\n", fast_mod(n));

    }
    return 0;
}

代码二： 用结构体封装矩阵乘法

#include<iostream>
#include<cstdio>
using namespace std;
const int MOD = 10000;
struct matrix
{
    int m[2][2];
}ans, base;
matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = 0; i < 2; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            tmp.m[i][j] = 0;
            for(int k = 0; k < 2; k++)
                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
        }
    }
    return tmp;
}
int fast_mod(int n)
{
    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
    base.m[1][1] = 0;
    ans.m[0][0] = ans.m[1][1] = 1;
    ans.m[0][1] = ans.m[1][0] = 0;
    while(n)
    {
        if(n & 1)
        {
            ans = multi(ans, base);
        }
        base = multi(base, base);
        n >>= 1;
    }
    return ans.m[0][1];
}
int main()
{
    int n;
    while(scanf("%d", &n) != EOF && n != -1)
    {
        printf("%d\n", fast_mod(n));
    }
    return 0;
}