Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 11587 | Accepted: 8229 |
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
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:
.
题意:就是求斐波那契数列,只是有点大,递推也会超时,矩阵快速幂0ms,搞定;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 using namespace std; 6 const int N = 2; 7 const int mod = 10000; 8 struct Mat 9 { 10 int mat[2][2]; 11 }; 12 Mat operator * (Mat a, Mat b) 13 { 14 Mat c; 15 memset(c.mat, 0, sizeof(c.mat)); 16 for(int k = 0; k < N; k++) 17 { 18 for(int i = 0; i < N; i++) 19 { 20 for(int j = 0; j < N; j++) 21 { 22 c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] % mod) % mod; 23 } 24 } 25 } 26 return c; 27 } 28 Mat operator ^ (Mat a, int k) 29 { 30 Mat c; 31 for(int i = 0; i < N; i++) 32 for(int j = 0; j < N; j++) 33 c.mat[i][j] = (i == j); 34 while(k) 35 { 36 if(k & 1) 37 c = c * a; 38 a = a * a; 39 k >>= 1; 40 } 41 return c; 42 } 43 int main() 44 { 45 int n; 46 while(scanf("%d", &n) != EOF) 47 { 48 if(n == -1) 49 break; 50 if(n == 0) 51 printf("0 "); 52 else if(n == 1) 53 printf("1 "); 54 else 55 { 56 Mat a; 57 a.mat[0][0] = 1; 58 a.mat[0][1] = 1; 59 a.mat[1][0] = 1; 60 a.mat[1][1] = 0; 61 a = a ^ (n - 1); 62 printf("%d ",a.mat[0][0]); 63 } 64 } 65 66 return 0; 67 }