| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12457 | Accepted: 8851 |
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:
.
Source
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 #include<queue> 5 #include<stack> 6 #include<map> 7 #define mod 10000 8 using namespace std; 9 struct matrix 10 { 11 int m[5][5]; 12 } ans,exm; 13 14 struct matrix matrix_mulit(struct matrix aa,struct matrix bb) 15 { 16 struct matrix there; 17 for(int i=0;i<2;i++) 18 { 19 for(int j=0;j<2;j++) 20 { 21 there.m[i][j]=0; 22 for(int k=0;k<2;k++) 23 there.m[i][j]=(there.m[i][j]+aa.m[i][k]*bb.m[k][j]%mod)%mod; 24 } 25 } 26 return there; 27 } 28 int matrix_quick(int gg) 29 { 30 exm.m[0][0]=exm.m[0][1]=exm.m[1][0]=1; 31 exm.m[1][1]=0; 32 ans.m[0][0]=ans.m[1][1]=1; 33 ans.m[0][1]=ans.m[1][0]= 0; 34 while(gg) 35 { 36 if(gg&1) 37 { 38 ans=matrix_mulit(ans,exm); 39 } 40 exm = matrix_mulit(exm, exm); 41 gg >>= 1; 42 } 43 return ans.m[0][0]; 44 } 45 int n; 46 int main() 47 { 48 while(scanf("%d",&n)!=EOF) 49 { 50 if(n==-1) 51 break; 52 printf("%d ",matrix_quick(n)); 53 } 54 return 0; 55 }