【2017沈阳网络赛】1005 hdu6198 number number number 找规律+矩阵快速幂

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

 

题意：

给出一个数k，问用k个斐波那契数相加，得不到的数最小是几。

思路：

找规律发现，答案就是f(2*k+3)-1，用矩阵快速幂求即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
typedef long long LL;
const LL Mod = 998244353;
class Matrix{    //阶数为n
public:
LL p[2][2];
Matrix(){
memset(p,0,sizeof(p));
p[0][0]=1;
p[1][1]=1;
}
Matrix mul(Matrix x,Matrix y){
Matrix c;
memset(c.p,0,sizeof(c.p));
for(int i=0;i<2;i++)
for(int j=0;j<2;j++){
for(int k=0;k<2;k++){
c.p[i][j]+=(x.p[i][k]*y.p[k][j])%Mod;
c.p[i][j]%=Mod;
}
}
return c;
}
Matrix pow(Matrix x,LL n){
Matrix ans,p=x;
while(n){
if(n&1)
ans=ans.mul(ans,p);
p=p.mul(p,p);
n>>=1;
}
return ans;
}
};
int main(int argc,const char*argv[]){
LL k;
while(scanf("%lld",&k)!=EOF){
LL n=2*k+3;
Matrix x;
Matrix y;
memset(x.p,0,sizeof(x.p));
memset(y.p,0,sizeof(y.p));
x.p[0][0]=1;
x.p[0][1]=1;
x.p[1][0]=1;
y.p[0][0]=1;
y.p[1][0]=0;
x=x.pow(x,n-1);
x=x.mul(x,y);
cout<<x.p[0][0]-1<<endl;
}
return 0;
}