So happy!(get 矩阵快速幂）ICPC :E HDU 6198

最新推荐文章于 2020-10-27 17:40:50 发布

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#icpc #数论 #矩阵快速幂 #斐波那契数列

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本文介绍了一种利用矩阵快速幂求解最小mjf-bad数的方法，通过观察斐波那契数列中bad数的规律，推导出bad数的计算公式，并给出具体实现代码。

number number number
Time Limit: 2000/1000 MS (Java/Others) Memory
Limit: 32768/32768 K (Java/Others) Total Submission(s): 637

Accepted Submission(s): 407
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
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
Sample Input

1

Sample Output

4
Source 2017 ACM/ICPC Asia Regional Shenyang Online

前提：矩阵快速幂+斐波那契数列推荐题目（poj3070（斐波那契的矩阵表示））
思路：
* 题目要求找bad number，枚举钱四项分别是 4,12,33,88,斐波那契数列为0 , 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89*
对应的下标分别是0,1,2,3,4,5,6,7,8,9,10,11
第一个bad number： 4=f4+f3-1；
第二个bad number ：12=f5+f6-1;
第三个bad number： 33 =f7+f8-1；
* 因而推导出一下公式： a(n)=f(2*n+2)+f(2*n+1)-1*

代码如下：

#include <bits/stdc++.h>
using namespace std;
const int mod =998244353;
struct matrix
{
   long long  int a[2][2];
};
matrix multply(matrix x,matrix y)
{
    matrix result ;
    memset(result.a,0,sizeof(result.a));
    int i,j,k;
    for (i=0;i<2;i++)
    for (j=0;j<2;j++)
    for (k=0;k<2;k++)
    {
        result.a[i][j]+=x.a[i][k]*y.a[k][j];
        if(result.a[i][j]>=mod)
        result.a[i][j]%=mod;
    }
    return result;
}
matrix pow_mod (matrix a,int n)
{
    matrix result;
    memset(result.a,0,sizeof(result.a));
    for (int i=0;i<2;i++)
    result.a[i][i]=1;
    while (n)
    {
        if(n&1)
         result=multply(result,a);
         a=multply(a,a);
         n>>=1;
    }
    return result ;

}

int main ()
{
    long long int n;
    while (cin>>n)
    {
         n=2*n+1;
         long long  int ans;
         matrix a;
         a.a[0][0]=a.a[0][1]=a.a[1][0]=1;
         a.a[1][1]=0;
         a=pow_mod(a,n);
         ans =a.a[0][0]+a.a[0][1]-1;
//         int i,j;
//         for (i=0;i<2;i++)
//         for (j=0;j<2;j++)
//         if(!j)
//         cout<<a.a[i][j]<<" ";
//         else
//         cout <<a.a[i][j]<<endl;
         if(ans>=mod)
         ans%=mod;
         cout<<ans<<endl;
    }
    return 0;
}