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

6
196

 

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
=s[n-1]+A
^2;A
^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
*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 |
*/