MIPT-2016 Pre-Finals Workshop, Taiwan NTU Contest Fibonacci of Fibonacci

Fibonacci of Fibonacci
题目大意：求第n项斐波那契数的斐波那契，即FFn mod 20160519
先找Fn mod 20160519的循环节,啊这个可以用板子求，然后用矩阵快速幂求一下就好了
 
 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 typedef long long ll;
 5 const int p = 20160519;
 6 const ll mode = 26880696;//用板子求的，有求斐波那契循环节的题目
 7 using namespace std;
 8 ll mod;
 9 struct mat {
10     ll a[3][3];
11     mat() {
12         memset(a, 0, sizeof(a));
13     }
14     mat operator *(const mat &o) const {
15         mat t;
16         for(int i = 1; i <= 2; i++) {
17             for(int j = 1; j <= 2; j++)
18                 for(int k = 1; k <= 2; k++) {
19                     t.a[i][j] = (t.a[i][j] + a[i][k]*o.a[k][j])%mod;
20                 }
21         }
22         return t;
23     }
24 } a, b;
25 ll slove(ll n) {
26     a.a[1][1] = a.a[2][2] = 1, a.a[1][2] = a.a[2][1] = 0;
27     b.a[1][1] = b.a[1][2] = b.a[2][1] = 1, b.a[2][2] = 0;
28     if(n == 0) return 0;
29     else {
30         n--;
31         while(n > 0) {
32             if(n&1) {
33                 a = b*a;
34                 n--;
35             }
36             n >>= 1;
37             b = b*b;
38         }
39        return a.a[1][1];
40     }
41 }
42 
43 int main() {
44     int t;
45     ll n;
46     scanf("%d", &t);
47     while(t--) {
48         scanf("%lld", &n);
49         mod = mode;
50         ll ans = slove(n);
51         mod = p;
52         ans = slove(ans);
53         printf("%lld\n", ans);
54     }
55     return 0;
56 }

 

 
 
 

 

posted on 2016-10-05 22:05  disppr  阅读(193)  评论(0)  编辑  收藏  举报