HDU6198-number number number

时间 2020-08-13 标签 hdu6198 hdu 6198 number number number

number number number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 619 Accepted Submission(s): 395

Problem Description

We define a sequence

⋅

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：给你一个斐波那契数列，问用其中k个数字最小的组不成的数字是几php

解题思路：多写几个后能够发现规律，斐波那契数列为0 1 1 2 3 5 8 13 21 34 55 89...k＝1时答案是4，k＝2时答案是12，k=3时答案是33，因此答案是第5.7.9.11....项-1ios

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const LL mod=998244353;

struct Matrix
{
    LL v[5][5];
    Matrix()
    {
        memset(v,0,sizeof v);
    }
} dan;

Matrix mul(Matrix a,Matrix b,int d)
{
    Matrix ans;
    for(int i=0; i<d; i++)
    {
        for(int j=0; j<d; j++)
        {
            for(int k=0; k<d; k++)
            {
                ans.v[i][j]+=(a.v[i][k]*b.v[k][j])%mod;
                ans.v[i][j]%=mod;
            }
        }
    }
    return ans;
}

Matrix pow(Matrix a,int k,int d)
{
    Matrix ans=dan;
    while(k)
    {
        if(k&1) ans=mul(ans,a,d);
        k>>=1;
        a=mul(a,a,d);
    }
    return ans;
}

int main()
{
    int k;
    while(~scanf("%d",&k))
    {
        dan.v[0][0]=1,dan.v[0][1]=0;
        Matrix a;
        a.v[0][0]=a.v[0][1]=a.v[1][0]=1,a.v[1][1]=0;
        Matrix ans= pow(a,2*k+2,2);
        printf("%lld\n",((ans.v[0][0]-1)%mod+mod)%mod);
    }
    return 0;
}