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

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

 

Statistic |  Submit |  Discuss |  Note
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);
    }
}