HDU - 6198 number number number(規律+矩陣快速冪)

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

1

 

Sample Output

4

 

Source

2017 ACM/ICPC Asia Regional Shenyang Online

 

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liuyiding

 

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題意：輸入一個 k ，輸出最小的不能由 k 個斐波那契數相加得到的正整數 n；
這題打表找規律，k = 1時 n = 4， k = 2時 n = 12，k = 3時 n = 33 ……規律即斐波那契數列的第 4+2*(k-1)項-1，最後用矩陣快速冪加速計算即可

#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);
    }
}