HDU3306 -- Another kind of Fibonacci 构造矩阵然后矩阵快速幂

Another kind of Fibonacci

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2428    Accepted Submission(s): 963

Problem Description

As we all known , the Fibonacci series : F(0) = 1, F(1) = 1, F(N) = F(N - 1) + F(N - 2) (N >= 2).Now we define another kind of Fibonacci : A(0) = 1 , A(1) = 1 , A(N) = X * A(N - 1) + Y * A(N - 2) (N >= 2).And we want to Calculate S(N) , S(N) = A(0)² +A(1)²+……+A(n)².

 

Input

There are several test cases.
Each test case will contain three integers , N, X , Y .
N : 2<= N <= 2³¹ – 1
X : 2<= X <= 2³¹– 1
Y : 2<= Y <= 2³¹ – 1

 

Output

For each test case , output the answer of S(n).If the answer is too big , divide it by 10007 and give me the reminder.

 

Sample Input

2 1 1 3 2 3 

 

Sample Output

6196

 

Author

wyb

 

Source

HDOJ Monthly Contest – 2010.02.06

 

Recommend

wxl

此题关键就在于构造矩阵，构造的矩阵如下：
很显然我们先推s[n]=s[n-1]+A[n]^2;A[n]^2=x*x*A[n-1]^2+y*y*A[n-2]^2+2*x*y*A[n-1][n-2];于是我们发现又涉及到了A[n-1]*A[n-2]。A[n]*A[n-1]=x*A[n-1]^2+y*A[n-1]*A[n-2]；

|2  1  1  1|   |1         0         0   0|  =  |f(n)       A(n)^2      A(n)*A(n-1)     A(n-1)^2|
|0  0  0  0| * |x^2     x^2      x   1|  =  |  0        0               0                     0            |
|0  0  0  0|   |2*x*y   2*x*y   y   0|  =  |  0        0               0                     0            |
|0  0  0  0|   |y^2     y^2      0   0|  =  |  0        0               0                     0            |

AC代码：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>using namespace std;const int maxm = 100002;const int maxn = 100002;const int inf = 0x3f3f3f3f;int n, m, k;struct node {int x, y, w;};node edge[maxm*2];int book[maxn];int main(){while(~scanf("%d%d%d", &n, &m, &k)) {memset(book, 0, sizeof(book));int i;for(i=0; i<m; ++i)scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].w);if(!k) {printf("-1\n");continue;}int a;for(i=0; i<k; ++i) {scanf("%d", &a);book[a] = 1;}int ans = inf;for(i=0; i<m; ++i) {if(book[edge[i].x] + book[edge[i].y] == 1) {ans = min(ans, edge[i].w);}}if(ans == inf)printf("-1\n");else  printf("%d\n", ans);}}/*|2  1  1  1|   |1       0        0   0| = |f(n)  A(n)^2  A(n)*A(n-1)  A(n-1)^2||0  0  0  0| * |x^2     x^2      x   1| = |  0     0         0            0   ||0  0  0  0|   |2*x*y   2*x*y    y   0| = |  0     0         0            0   ||0  0  0  0|   |y^2     y^2      0   0| = |  0     0         0            0   |*/