HDU - 6198 number number number(规律+矩阵快速幂)

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探讨了如何通过矩阵快速幂算法找到对于给定k值的最小mjf-bad数，该数无法由k个斐波那契数之和表示，并提供了一种高效的计算方法。

number number number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 274 Accepted Submission(s): 174

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

Sample Output

Source

2017 ACM/ICPC Asia Regional Shenyang Online

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题意：输入一个 k ，输出最小的不能由 k 个斐波那契数相加得到的正整数 n；
这题打表找规律，k = 1时 n = 4， k = 2时 n = 12，k = 3时 n = 33 ……规律即斐波那契数列的第 4+2*(k-1)项-1，最后用矩阵快速幂加速计算即可

#include <bits/stdc++.h>
using namespace std;

const long long N = 2, Mod = 998244353;
long long m, n;
struct xx{
    long long a[N][N];
} ori,res;

xx mul(xx x,xx y){
    xx temp;
    memset(temp.a, 0, sizeof(temp.a));
    for(int i = 0; i < N; i++){
        for(int j = 0; j < N; j++){
            for(int k = 0; k < N; k++){
                temp.a[i][j] += x.a[i][k]*y.a[k][j];
                temp.a[i][j] %= Mod;
            }
        }
    }
    return temp;
}

void init(){
    ori.a[0][0] = 1, ori.a[0][1] = 1;
    ori.a[1][0] = 1, ori.a[1][1] = 0;
    res.a[0][0] = 1, res.a[0][1] = 0;
    res.a[1][0] = 0, res.a[1][1] = 1;
}

long long calc(long long k){
    while(k){
        if(k & 1) res = mul(res, ori);
        ori = mul(ori, ori);
        k >>= 1;
    }
    return res.a[0][0];
}

int main(){
    while(scanf("%lld", &m) == 1){
        init();
        n = 4+(m-1)*2;
        printf("%lld\n", calc(n)-1);
    }
}