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HDU6198

时间:2017-09-10 21:42:23      阅读:172      评论:0      收藏:0      [点我收藏+]

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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 (n2).

Give you an integer k, if a positive number n can be expressed by
n=Fa1+Fa2+...+Fak where 0a1a2?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. (1k109)
 

 

Output

For each case, output the minimal mjf?bad number mod 998244353.
 

 

Sample Input

1
 

 

Sample Output

4
 

 

Source

 
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 }

 

HDU6198

标签:title   express   min   preview   include   text   each   tin   osi   

原文地址:http://www.cnblogs.com/Penn000/p/7502280.html

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