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自然万物都趋向从有序变得无序

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number number number hdu 6189 矩阵快速幂

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题目链接

题目

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

Source

2017 ACM/ICPC Asia Regional Shenyang Online

题意:

给你一个 k, 计算不能用 k 个可以重复的斐波那契数列组成的最小的数.

思路:

打表找规律, 我们发现k = 1 时, 答案为 F5−1F_5 - 1, (F为斐波那契数列), k = 2 时, 答案为 F7−1F_7 - 1, k = 3 时, 答案为 F9−1F_9 - 1, 验证一下发现k = 4 时答案也成立, 然后就是矩阵快速幂,

其实我觉得是可以证明的, 但是鶸表示不会, 我的猜测大概思路应该是因为k = 1 时候, 答案为 4, 然后k = 2 时的答案一定由4 加上某个数组成, 然后要使这个数最小, 所以找一个比 4 大, 而且不为 5 的最小的斐波那契数, 所以 k = 2的答案应该是 4 + 8, 就等于F7−1F_7 - 1 依次类推,—博主不敢保证思路正确, 若有人会证明请告诉博主.

代码:

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;

typedef long long LL;
const int mod = 998244353;
LL k;

struct mat {
    LL a[3][3];
    mat(){memset(a, 0,sizeof(a)); }
    mat operator *(const mat q){
        mat c;
        for(int i = 1; i <= 2; ++i)
            for(int j = 1; j <= 2; ++j)
            if(a[i][j])
        for(int k = 1;  k<= 2; ++k){
            c.a[i][k] += a[i][j] * q.a[j][k];
            c.a[i][k] %= mod;
        }return c;
    }
};

mat qpow(mat x, LL n){
    mat ans;
    ans.a[1][1] = ans.a[2][2] = 1;
    while(n){
        if(n&1) ans = ans * x;
        x = x * x;
        n >>= 1;
    }return ans;
}

LL qpow(LL x,LL n){
    LL ans = 1;
    while(n){
        if (n&1) ans = (ans * x) %( mod + 1);
        x = (x * x) % (mod + 1);
        n >>= 1;
    }return ans;
}

int main(){
    mat tmp, ans;
    tmp.a[1][1] = tmp.a[1][2] = tmp.a[2][1] = 1;
    while(scanf("%lld", &k) != EOF){
        ans = qpow(tmp, 2 * k + 3);
        printf("%lld\n",(((ans.a[2][1] -1) % mod) + mod) % mod);
    }return 0;
}