hdu 4686 Arc of Dream

题目链接： 

　　http://acm.hdu.edu.cn/showproblem.php?pid=4686

题目描述：

　　题目上说的很清楚。

解题思路：

　　就是递推，用快速幂+构造矩阵解决，因为n的取值范围好大，好全。

/*
    |1 0      0   0   0 |
    |1 ax*bx  0   0   0 |
    |0 ax*by  ax  0   0 |*|s[n] f[n] a[n] b[n] 1|=|s[n+1] f[n+1] a[n+1] b[n+1] 1|
    |0 bx*ay  0   bx  0 |
    |0 by*ay  ay  by  1 |
    
    a[n+1]=a[n]*ax+ay;
    b[n+1]=b[n]*bx+by;
    f[n]=a[n]*b[n];
    s[n+1]=s[n]+f[n];
*/
 1 #include <iostream>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <queue>
 7 #include <cstring>
 8 using namespace std;
 9 
10 const int maxn = 5;
11 #define LL long long
12 #define mod 1000000007
13 struct mat
14 {
15     LL p[maxn][maxn];
16 };
17 
18 mat mul (mat a, mat b);
19 mat pow (LL n, mat a, mat b);
20 int main ()
21 {
22     LL n, a, ax, ay, b, bx, by;
23     mat A, B;
24     while (scanf ("%lld", &n) != EOF)
25     {
26         memset (A.p, 0, sizeof(A.p));
27         memset (B.p, 0, sizeof(B.p));
28         scanf ("%lld %lld %lld %lld %lld %lld", &a, &ax, &ay, &b, &bx, &by);
29         a %= mod, b %= mod, ax %= mod, bx %= mod, ay %= mod, by %= mod;
30         B.p[0][0] = 0;//贴这个代码，我什么也不想说明，只想表明矩阵相乘取余要全面，要细心，（wa哭了）
31         B.p[0][1] = a*b%mod;//心好累，在上面取完模还不够，在这里也要取模
32         A.p[1][1] = ax*bx%mod;
33         A.p[2][1] = ax*by%mod;
34         A.p[3][1] = ay*bx%mod;
35         A.p[4][1] = ay*by%mod;
36         B.p[0][2] = a;
37         B.p[0][3] = b;
38         B.p[0][4] = 1;
39         A.p[0][0] = A.p[1][0] = 1;
40         
41         A.p[2][2] = ax;
42         A.p[3][3] = bx;
43         A.p[4][4] = 1;
44         A.p[4][2] = ay;
45         A.p[4][3] = by;
46         B = pow (n, A, B);
47         printf ("%lld\n", B.p[0][0]);
48     }
49     return 0;
50 }
51 
52 mat mul (mat a, mat b)
53 {
54     int i, j, k;
55     mat c;
56     memset (c.p, 0, sizeof(c.p));
57 
58     for (i=0; i<5; i++)
59         for (j=0; j<5; j++)
60         {
61             for (k=0; k<5; k++)
62                 c.p[i][j] = (c.p[i][j] + a.p[i][k] * b.p[k][j]) % mod;
63         }
64     return c;
65 }
66 mat pow (LL n, mat a, mat b)
67 {
68     while (n)
69     {
70         if (n % 2)
71         {
72             b = mul (b, a);
73         }
74         a = mul (a, a);
75         n /= 2;
76     }
77     return b;
78 }

 

posted @ 2015-03-29 20:56  罗茜  阅读(130)  评论(0)  编辑  收藏  举报