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最新推荐文章于 2019-05-20 18:33:28 发布

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本文介绍了一种使用矩阵快速幂方法来寻找最小的mjf-bad正整数的问题解决思路。通过观察斐波那契数列的性质，文章提供了一个高效的算法实现，并给出了完整的C++代码示例。

Problem Description
We define a sequence F:

⋅ F0=0,F1=1;
⋅ Fn=Fn−1+Fn−2 (n≥2).

Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+…+Fak where 0≤a1≤a2≤⋯≤ak, this positive number is mjf−good. Otherwise, this positive number is mjf−bad.
Now, give you an integer k, you task is to find the minimal positive mjf−bad number.
The answer may be too large. Please print the answer modulo 998244353.

Input
There are about 500 test cases, end up with EOF.
Each test case includes an integer k which is described above. (1≤k≤109)

Output
For each case, output the minimal mjf−bad number mod 998244353.

Sample Input
1

Sample Output
4
先写出几项找到规律是求第i个mjf−bad number就是第2*i+3项斐波那契数列的值。
矩阵快速幂模版题。之前一直没做矩阵快速幂的题，还好队友写出来了。

#include <iostream>
#include <cstdio>
#include <cstring>
#define LL long long
#define MOD 998244353

using namespace std;

struct matrix
{
    LL x[3][3];
};

struct matrix mutimatrix(struct matrix a,struct matrix b)
{
    struct matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    for(int i=0;i<2;i++)
    {
        for(int j=0;j<2;j++)
        {
            for(int k=0;k<2;k++)
            {
                temp.x[i][j]=temp.x[i][j]+a.x[i][k]*b.x[k][j];
                temp.x[i][j]=temp.x[i][j]%MOD;
            }
        }
    }
    return temp;
}

struct matrix pow_matrix(struct matrix a,LL b)
{
    struct matrix temp;
    memset(temp.x,0,sizeof(temp.x));
    for(int i=0;i<2;i++)
    {
        temp.x[i][i]=1;
    }
    while(b)
    {
        if(b&1)
        {
            temp=mutimatrix(temp,a);
        }
        a=mutimatrix(a,a);
        b>>=1;
    }
    return temp;
}

int main()
{
    LL n;
    while(scanf("%lld",&n)!=EOF)
    {
        struct matrix temp,ans;
        temp.x[0][0]=1;
        temp.x[0][1]=1;
        temp.x[1][0]=1;
        temp.x[1][1]=0;
        ans=pow_matrix(temp,2*n+3);
        printf("%lld\n",ans.x[0][1]-1);
    }
    return 0;
}