Hdu--4990（矩阵快速幂，模板）

2014-09-08 00:26:02

贴一发矩阵快速幂的模板。

 1 /*************************************************************************
 2     > File Name: b.cpp
 3     > Author: Nature
 4     > Mail: 564374850@qq.com
 5     > Created Time: Sun 07 Sep 2014 10:50:06 PM CST
 6 ************************************************************************/
 7 
 8 #include <cstdio>
 9 #include <cstring>
10 #include <cstdlib>
11 #include <cmath>
12 #include <iostream>
13 #include <algorithm>
14 using namespace std;
15 typedef long long ll;
16 const int INF = 1 << 29;
17 
18 int n,m;
19 
20 struct Mx{
21     int a[2][2];
22     void clear(){ a[0][0] = a[0][1] = a[1][0] = a[1][1] = 0;}
23     void stand(){ a[0][0] = a[1][1] = 1,a[0][1] = a[1][0] = 0;}
24     Mx operator * (const Mx &b) const{
25         Mx c;
26         c.clear();
27         for(int k = 0; k < 2; ++k)
28         for(int i = 0; i < 2; ++i)
29         for(int j = 0; j < 2; ++j){
30             c.a[i][j] = (c.a[i][j] + (ll)a[i][k] * b.a[k][j]) % m;
31         }
32         return c;
33     }
34 };
35 
36 ll Mx_pow(int n){
37     Mx y,a;
38     y.stand();
39     a.a[0][0] = 4,a.a[0][1] = a.a[1][1] = 1,a.a[1][0] = 0;
40     while(n){
41         if(n & 1) y = y * a;
42         if(n >>= 1) a = a * a;
43     }
44     return y.a[0][1] % m;
45 }
46 
47 int main(){
48     while(scanf("%d%d",&n,&m) != EOF){
49         int ans = Mx_pow((n + 1) / 2);
50         if((n & 1) == 0) ans *= 2;
51         printf("%d\n",ans % m);
52     }
53     return 0;
54 }

 

posted @ 2014-09-08 00:26  Naturain  阅读(167)  评论(0)  编辑  收藏  举报