poj 3070 Fibonacci (矩阵加速线性递推式)
题意：求f[n] % 10000,f为斐波那契数。
思路：按照题目给出的公式，或者按照加速线性递推式的方法都可以。。。
因为把模数的1E4手滑写成1E4+7结果调了半天也是没谁了呵呵呵呵。
/* ***********************************************
Author :111qqz
Created Time :Tue 18 Oct 2016 01:18:40 AM CST
File Name :code/poj/3070.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int MOD = 1E4;
7int n;
8struct Mat
9{
10 int mat[5][5];
1 void clear()
2 {
3 ms(mat,0);
4 }
1}M,M1;
2Mat operator * (Mat a,Mat b)
3{
4 Mat c;
5 c.clear();
6 for ( int i = 0 ; i < 2 ; i++)
7 for ( int j = 0 ; j < 2 ; j++)
8 for ( int k = 0 ; k < 2 ; k++)
9 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k]*b.mat[k][j])%MOD;
10 return c;
11}
12Mat operator ^ (Mat a,int b)
13{
14 Mat c;
15 c.clear();
16 for ( int i = 0 ; i < 2 ; i++)
17 for ( int j = 0 ; j < 2 ; j++)
18 c.mat[i][j]=(i==j); //初始化为单位矩阵。
19 while (b>0)
20 {
21 if (b&1) c = c * a;
22 b = b >> 1;
23 a = a * a;
24 }
25 return c;
26}
27int main()
28{
29 #ifndef ONLINE_JUDGE
30 freopen("code/in.txt","r",stdin);
31 #endif
1 while (~scanf("%d",&n))
2 {
3 if (n==-1) break;
4 if (n==0)
5 {
6 printf("%d\n",0);
7 continue;
8 }
9 M.mat[0][0] = 0;
10 M.mat[0][1] = 1;
11 M.mat[1][0] = 1;
12 M.mat[1][1] = 1;
13 M1.mat[0][0] = 0;
14 M1.mat[1][0] = 1;
15 Mat ans;
16 ans.clear();
17 ans = (M ^ n )* M1;
18 printf("%d\n",ans.mat[0][0]);
19 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}