hdu4686 Arc of Dream ——构造矩阵+快速幂

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

构造出来的矩阵是这样的：根据题目的ai * bi = ……，可以发现 矩阵1 * 矩阵3 = 矩阵2。然后就是矩阵快速幂了。

1

1	ai	bi	ai*bi	Si
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0

2

1	ai+1	bi+1	ai+1*bi+1	Si+1
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0

3

1	AY	BY	AY*BY	AY*BY
0	AX	0	AX*BY	AX*BY
0	0	BX	AY*BX	AY*BX
0	0	0	AX*BX	AX*BX
0	0	0	0	1

 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #include <queue>
 #include <deque>
 #include <queue>
 #include <list>
 #include <map>
 #include <set>
 #include <vector>
 #include <utility>
 #include <functional>
 #include <fstream>
 #include <iomanip>
 #include <sstream>
 #include <numeric>
 #include <cassert>
 #include <ctime>
 #include <iterator>
 const int INF = 0x3f3f3f3f;
 ][] = {{-,},{,},{,-},{,},{-,-},{-,},{,-},{,}};
 using namespace std;
 #define LL __int64
 #define MOD 1000000007
 typedef struct
 {
     LL m[][];
 }mat;
 mat X, Y;
 LL n, a0, ax, ay, b0, bx, by;
 mat multi(mat a, mat b)
 {
     mat c; int j, i, k;
     ; i < ; ++i)
     {
         ; j < ; ++j)
         {
             c.m[i][j] = ;
             ; k < ; ++k)
             {
                 c.m[i][j] += a.m[i][k] * b.m[k][j]%MOD;
             }
             c.m[i][j] %= MOD;
         }
     }
     return c;
 }
 mat power(LL k)
 {
     mat ans = X, p = Y;
     while (k)
     {
         ) ans = multi(ans, p);
         k /= ; p = multi(p, p);
     }
     return ans;
 }
 int main(void)
 {
     #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin );
     #endif // ONLINE_JUDGE
     ios::sync_with_stdio(false);
     while (cin>>n>>a0>>ax>>ay>>b0>>bx>>by)
     {
         "<<endl; continue;}
         memset(X.m, , sizeof(X.m));
         memset(Y.m, , sizeof(Y.m));
         X.m[][] = , X.m[][] = a0%MOD, X.m[][] = b0%MOD,
         X.m[][] = a0*b0%MOD, X.m[][] = a0*b0%MOD;
         Y.m[][] = , Y.m[][] = ay%MOD, Y.m[][] = by%MOD,
         Y.m[][] = ay*by%MOD, Y.m[][] = ay*by%MOD, Y.m[][] = ax%MOD,
         Y.m[][] = ax*by%MOD, Y.m[][] = ax*by%MOD, Y.m[][] = bx%MOD,
         Y.m[][] = ay*bx%MOD, Y.m[][] = ay*bx%MOD, Y.m[][] = ax*bx%MOD,
         Y.m[][] = ax*bx%MOD, Y.m[][] = ;
         mat ans = power(n-);
         LL touch = ans.m[][];
         cout << touch <<endl;
     }
     ;
 }

注意 n==0 的时候特判呐~

走吧，小胖！