2013 Multi-University Training Contest 9 1011 Arc of Dream

利用递推关系构建矩阵 然后矩阵快速幂

 1 #include<cstdio>
 2 #include<cstring>
 3 typedef long long LL;
 4 const int mod=1000000007;
 5 struct matrix
 6 {
 7     LL m[5][5];
 8     matrix(){memset(m,0,sizeof(m));}
 9     matrix operator*(const matrix &a)const
10     {
11         matrix tmp;
12         for(int i=0;i<5;i++)
13             for(int j=0;j<5;j++)
14                for(int k=0;k<5;k++)
15                 tmp.m[i][j]=(tmp.m[i][j]+m[i][k]*a.m[k][j]%mod)%mod;
16         return tmp;
17     }
18 }e;
19 matrix pow(matrix a,LL n)
20 {
21    if(n==1) return  a;
22    if(n%2)
23    {
24       matrix tmp=pow(a,n/2);
25       return tmp*tmp*a;
26    }
27    else
28    {
29        matrix tmp=pow(a,n/2);
30        return tmp*tmp;
31    }
32 }
33 int main()
34 {
35     LL n;
36     while(scanf("%lld",&n)!=EOF)
37     {
38         LL A0,AX,AY,B0,BX,BY;
39         scanf("%lld%lld%lld%lld%lld%lld",&A0,&AX,&AY,&B0,&BX,&BY);
40          if(n==0) {printf("0\n");continue;}
41         LL input[5][5]={AX*BX%mod,AX*BY%mod,AY*BX%mod,AY*BY%mod,0,0,AX%mod,0,AY%mod,0,0,0,BX%mod,BY%mod,0,0,0,0,1,0,AX*BX%mod,AX*BY%mod,AY*BX%mod,AY*BY%mod,1};
42         LL a[5]={A0*B0%mod,A0%mod,B0%mod,1,A0*B0%mod};
43         for(int i=0;i<5;i++)
44             for(int j=0;j<5;j++)
45              e.m[i][j]=input[i][j];
46         if(n==1) { printf("%lld\n",a[4]);continue;}
47         matrix ans=pow(e,n-1);
48         LL  answer=0;
49         for(int i=0;i<5;i++)
50             answer=(answer+ans.m[4][i]*a[i]%mod)%mod;
51         printf("%lld\n",answer);
52     }
53     return 0;
54 }

 

posted @ 2013-08-21 15:02  sooflow  阅读(190)  评论(0编辑  收藏  举报