矩阵快速幂

来源:互联网 发布:数据库毕业论文题目 编辑:程序博客网 时间:2024/06/06 00:41

对于n比较大的斐波那契

#include <cstdio>
#include <iostream>
using namespace std;
const long long int MOD =1000000007;
struct matrix
{
    long long 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;
}
long long int fast_mod(long long 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] = 1;
    while(n)
    {
        if(n & 1)
            ans = multi(ans, base);
        base = multi(base, base);
        n >>= 1;
    }
    return ans.m[0][1];
}
int main()
{
   long long int n;
    while(scanf("%lld", &n))
    {
        if(n==0)
            break;
        printf("%lld\n", fast_mod(n));
    }
    return 0;
}

0 0
原创粉丝点击