POJ 3070(求斐波那契数 矩阵快速幂)

题意就是求第 n 个斐波那契数。

由于时间和内存限制，显然不能直接暴力解或者打表，想到用矩阵快速幂的做法。

代码如下：

 1 #include <cstdio>
 2 using namespace std;
 3 const int maxn = 100;
 4 const int mod = 10000;
 5 int a;
 6 struct Matrix
 7 {
 8     int m[maxn][maxn];
 9 }ans,res,w,head;
10 
11 Matrix mul(Matrix a,Matrix b,int n)
12 {
13     Matrix tmp;
14     for(int i = 1; i <= n; i++)
15         for(int j = 1; j <= n; j++)
16             tmp.m[i][j] = 0;
17     for(int i = 1; i <= n; i++)
18         for(int j = 1; j <= n; j++)
19             for(int k = 1; k <= n; k++)
20                 tmp.m[i][j] += ((a.m[i][k] % mod)*(b.m[k][j] % mod))%mod;
21     return tmp;
22 }
23 
24 void quickpow(int N,int n)
25 {
26     for(int i = 1; i <= n; i++)
27         for(int j = 1; j <= n; j++)
28             if(i == j)  ans.m[i][j] = 1;
29             else ans.m[i][j] = 0;
30     while(N)
31     {
32         if(N&1) ans = mul(ans,res,2);
33         res = mul(res,res,2);
34         N = N >>1;
35     }
36 }
37 
38 int main()
39 {
40     while(scanf("%d",&a))
41     {
42         if(a == -1) break;
43         else if(a == 0)
44         {
45             puts("0");
46             continue;
47         }
48         head.m[1][1] = head.m[1][2] = head.m[2][2] = 0;
49         head.m[2][1] = 1;
50         res.m[1][1] = res.m[1][2] = res.m[2][1] = 1;
51         res.m[2][2] = 0;
52         quickpow(a,2);
53         w = mul(ans,head,2);
54         printf("%d\n",w.m[1][1]);
55     }
56     return 0;
57 }
View Code

 

posted @ 2018-08-14 22:03  Taskr  阅读(139)  评论(0)  编辑  收藏  举报
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