HDU6198

Posted 2020-10-06 Penn000

number number number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 118    Accepted Submission(s): 79

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

 

ans = F(2*n+1)-1

 

 1 //2017-09-10
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 #define LL long long
 7 #define MAXN 100
 8 
 9 using namespace std;
10 
11 const int MOD = 998244353;
12 
13 struct Matrix  
14 {  
15     LL a[MAXN][MAXN];  
16     int r, c; 
17 };  
18 
19 Matrix ori, res; 
20 
21 void init()  
22 {  
23     memset(res.a, 0, sizeof(res.a));  
24     res.r = 2; res.c = 2;  
25     for(int i = 1; i <= 2; i++)  
26       res.a[i][i] = 1;  
27     ori.r = 2; ori.c = 2;  
28     ori.a[1][1] = ori.a[1][2] = ori.a[2][1] = 1;  
29     ori.a[2][2] = 0;  
30 }  
31 
32 Matrix multi(Matrix x, Matrix y)  
33 {  
34     Matrix z;  
35     memset(z.a, 0, sizeof(z.a));  
36     z.r = x.r, z.c = y.c;    
37     for(int i = 1; i <= x.r; i++) 
38     {  
39         for(int k = 1; k <= x.c; k++)      
40         {  
41             if(x.a[i][k] == 0) continue;
42             for(int j = 1; j<= y.c; j++)  
43               z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;  
44         }  
45     }  
46     return z;  
47 }  
48 void Matrix_mod(int n)  
49 {  
50     while(n)  
51     {  
52         if(n & 1)  
53           res = multi(ori, res);  
54         ori = multi(ori, ori);  
55         n >>= 1;  
56     }  
57     printf("%lld\n", res.a[1][2]-1 % MOD);  
58 }  
59 
60 int main()
61 {
62     int k;
63     while(scanf("%d", &k) != EOF)
64     {
65         init();
66         k++;
67         Matrix_mod(2*k+1);
68     }
69     return 0;
70 }