hdu 2604 Queuing（矩阵快速幂）

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<iostream>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=15;
 7 typedef long long ll;
 8 ll n=7,mod;
 9 ll e[maxn][maxn],v[maxn][maxn],w[maxn][maxn];
10 void mul(ll a[maxn][maxn],ll b[maxn][maxn])
11 {
12 
13     memset(e,0,sizeof(e));
14     for(ll i=1;i<=n;++i)
15     {
16         for(ll j=1;j<=n;++j)
17         {
18             ll sum=0;
19             for(ll k=1;k<=n;++k)
20             {
21                 sum=(sum%mod+(a[i][k]*b[k][j])%mod)%mod;
22             }
23             e[i][j]=sum;
24         }
25     }
26     for(ll i=1;i<=n;++i)
27     {
28         for(ll j=1;j<=n;++j)
29         {
30             a[i][j]=e[i][j];
31         }
32     }
33 }
34 void pow(ll b)
35 {
36     for(ll i=1;i<=n;++i)
37     {
38         w[i][i]=1;
39     }
40     while(b)
41     {
42         if(b&1)
43         {
44             mul(w,v);
45         }
46         mul(v,v);
47         b>>=1;
48     }
49 }
50 int main()
51 {
52     ll a,b;
53     while(~scanf("%lld%lld",&a,&b))
54     {
55         mod = b;
56         memset(w,0,sizeof(w));
57         memset(v,0,sizeof(v));
58         if(a==1)
59         {
60             int c=2%mod;
61             printf("%d\n",c);
62             continue;
63         }
64         else if(a==2)
65         {
66             int c=4%mod;
67             printf("%d\n",c);
68             continue;
69         }
70         else if(a==3)
71         {
72             int c=6%mod;
73             printf("%d\n",c);
74             continue;
75         }
76         else if(a==4)
77         {
78             int c=9%mod;
79             printf("%d\n",c);
80             continue;
81         }
82         else if(a==0)
83         {
84             printf("0\n");
85             continue;
86         }
87         //0 2 4 6 9
88         v[1][1]=1;v[1][2]=0;v[1][3]=1;v[1][4]=1;
89         v[2][1]=1;v[2][2]=0;v[2][3]=0;v[2][4]=0;
90         v[3][1]=0;v[3][2]=1;v[3][3]=0;v[3][4]=0;
91         v[4][1]=0;v[4][2]=0;v[4][3]=1;v[4][4]=0;
92         pow(a-4);
93         ll sum=(w[1][1]*9+w[1][2]*6+w[1][3]*4+w[1][4]*2)%mod;
94         printf("%lld\n",sum);
95     }
96     return 0;
97 }