| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 7241 | Accepted: 5131 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.
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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
099999999991000000000-1
Sample Output
0346266875
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 #include2 #include 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}; // 初始化为单位矩阵12 int tmp[2][2]; //自始至终都作为矩阵乘法中的中间变量 13 14 while(n)15 {16 if(n & 1) //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 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; // 注意这里要都赋值成 0 22 23 for(int i = 0; i < 2; ++i) // 矩阵乘法 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 // 下边要实现 t *= t 的操作，同样要先将t赋值给中间变量 tmp ，t清零，之后 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 }
代码二：用结构体封装矩阵乘法后，代码看着清晰多了
1 #include2 #include 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) // 求矩阵 base 的 n 次幂 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 初始化为单位矩阵 32 ans.m[0][1] = ans.m[1][0] = 0;33 while(n)34 {35 if(n & 1) //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 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 }