Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 const int MOD = 10000;
 7 
 8 int fast_mod(int n)    // 求 (t^n)%MOD 
 9 {
10     int t[2][2] = {1, 1, 1, 0};
11     int ans[2][2] = {1, 0, 0, 1};  // 初始化為單位矩陣
12     int tmp[2][2];    //自始至終都作為矩陣乘法中的中間變數 
13      
14     while(n)
15     {
16         if(n & 1)  //實現 ans *= t; 其中要先把 ans賦值給 tmp，然後用 ans = tmp * t 
17         {
18             for(int i = 0; i < 2; ++i)
19                 for(int j = 0; j < 2; ++j)
20                     tmp[i][j] = ans[i][j]; 
21             ans[0][0] = ans[1][1] = ans[0][1] = ans[1][0] = 0;  // 注意這裡要都賦值成 0 
22             
23             for(int i = 0; i < 2; ++i)    //  矩陣乘法 
24             {
25                 for(int j = 0; j < 2; ++j)
26                 {
27                     for(int k = 0; k < 2; ++k)
28                         ans[i][j] = (ans[i][j] + tmp[i][k] * t[k][j]) % MOD;
29                 }
30             }
31         }
32         
33         //  下邊要實現  t *= t 的操作，同樣要先將t賦值給中間變數  tmp ，t清零，之後 t = tmp* tmp 
34         for(int i = 0; i < 2; ++i)
35             for(int j = 0; j < 2; ++j)
36                 tmp[i][j] = t[i][j];
37         t[0][0] = t[1][1] = 0;
38         t[0][1] = t[1][0] = 0;
39         for(int i = 0; i < 2; ++i)
40         {
41             for(int j = 0; j < 2; ++j)
42             {
43                 for(int k = 0; k < 2; ++k)
44                     t[i][j] = (t[i][j] + tmp[i][k] * tmp[k][j]) % MOD;
45             }
46         }
47         
48         n >>= 1;
49     }
50     return ans[0][1];
51 }
52 
53 int main()
54 {
55     int n;
56     while(scanf("%d", &n) && n != -1)
57     {    
58         printf("%d\n", fast_mod(n));
59     }
60     return 0;
61 }

程式碼二：用結構體封裝矩陣乘法後，程式碼看著清晰多了

 1 #include <cstdio>
 2 #include <iostream>
 3 
 4 using namespace std;
 5 
 6 const int MOD = 10000;
 7 
 8 struct matrix
 9 {
10     int m[2][2];
11 }ans, base;
12 
13 matrix multi(matrix a, matrix b)
14 {
15     matrix tmp;
16     for(int i = 0; i < 2; ++i)
17     {
18         for(int j = 0; j < 2; ++j)
19         {
20             tmp.m[i][j] = 0;
21             for(int k = 0; k < 2; ++k)
22                 tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
23         }
24     }
25     return tmp;
26 }
27 int fast_mod(int n)  // 求矩陣 base 的  n 次冪 
28 {
29     base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
30     base.m[1][1] = 0;
31     ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化為單位矩陣 
32     ans.m[0][1] = ans.m[1][0] = 0;
33     while(n)
34     {
35         if(n & 1)  //實現 ans *= t; 其中要先把 ans賦值給 tmp，然後用 ans = tmp * t 
36         {
37             ans = multi(ans, base);
38         }
39         base = multi(base, base);
40         n >>= 1;
41     }
42     return ans.m[0][1];
43 }
44 
45 int main()
46 {
47     int n;
48     while(scanf("%d", &n) && n != -1)
49     {   
50         printf("%d\n", fast_mod(n));
51     }
52     return 0;
53 }